To calculate the average acceleration between the times t=1.90 s and t=3.10 s, we need to find the change in velocity over this time interval. The position of a particle on the x-axis, according to
[tex]x(t)=11.2t−1.50t^2 m.[/tex]
The velocity of the particle at t=1.90 s is given by v(1.90) = 11.2 - 3(1.90) = 5.5 m/s
The velocity of the particle at t=3.10 s is given by v(3.10) = 11.2 - 3(3.10) = 1.9 m/s
The change in velocity over this time interval is given by Δv = v(3.10) - v(1.90)
Δv = 1.9 - 5.5 = -3.6 m/s
The average acceleration between the times t=1.90 s and t=3.10 s is given by:
avg = Δv / Δt
where Δt = 3.10 s - 1.90 s = 1.20 s
avg = -3.6 m/s / 1.20 s = -3 m/s²
Therefore, the average acceleration between the times t=1.90 s and t=3.10 s is -3 m/s².
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Adjust the Applied Force by moving the slider bar (which is at the center bottom of the main window) left or right. Release the slider to stop applying a force on the object. You can also adjust the Applied Force in increments of 50 Newtons by clicking on the left and right arrows above the slider bar. Apply a force on the crate and watch the resulting motion. Feel free to play around with the simulation, and investigate what happens when you click on either the figure or the object, while they are in motion. When you are done, hit reset (round button with a circular arrow, to the right) and set Friction to None prior to beginning Part A. Make sure that Forces, Sum of Forces, Values, Masses, Speed, and Acceleration are all selected.
Choose the 200 kg refrigerator. Set the applied force to 400 NN (to the right). Be sure friction is turned off.
What is the net force acting on the refrigerator?
Choose the 200 refrigerator. Set the applied force to 400 (to the right). Be sure friction is turned off.
What is the net force acting on the refrigerator?
The net force is zero.
The magnitude of the net force is 400N, directed to the right.
The magnitude of the net force is less than 400N, directed to the right.
The magnitude of the net force is greater than 400N, directed downward and to the right.
The magnitude of the net force is greater than 400N, directed upward and to the right.
When the applied force on the 200 kg refrigerator is set to 400 N to the right and friction is turned off, the net force acting on the refrigerator is 400 N to the right.
When the applied force on the 200 kg refrigerator is set to 400 N to the right, and friction is turned off, the net force acting on the refrigerator is 400 N, directed to the right. This means that the total force exerted on the refrigerator in the horizontal direction is 400 N.
The net force is calculated by considering all the forces acting on an object. In this case, there are no other forces involved apart from the applied force. Since the friction is turned off, there is no opposing force to counteract the applied force. As a result, the applied force becomes the net force acting on the refrigerator.
It's important to note that the magnitude of the net force is the same as the magnitude of the applied force, which is 400 N. The direction of the net force is determined by the direction of the applied force, which in this case is to the right.
Overall, when the applied force is set to 400 N to the right and friction is turned off, the net force acting on the 200 kg refrigerator is 400 N, directed to the right.
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How much will the length of a simple pendulum be if its time period is one second?
If the time period is one second the length of a simple pendulum will be is 0.25m .
The length is calculated by the time period formula
[tex]T = 2\pi \sqrt{\frac{l}{g} }[/tex]. . . . . . . .(1)
where T = time period
l = length
g = gravitational constant
As per the question
Time period = 1 second
Gravitational constant = [tex]10 m/s^{2}[/tex]
Putting the values in equation (1) we get
[tex]1 = 2\pi \sqrt{\frac{l}{10} }[/tex] . . . . . . . . (2)
As we know the value of [tex]\pi[/tex] is 3.14
Therefore substituting the value of [tex]\pi[/tex] in equation 2 we get
[tex]1 = 2 X 3.14\sqrt{\frac{l}{10} }\\1 = 6.28 \sqrt{\frac{l}{10} }[/tex]
Squaring both the sides we get
[tex]1 =39.43 X\frac{l}{10}[/tex]
[tex]l = \frac{10}{39.43}[/tex]
[tex]l= 0.25 m[/tex] ( approx )
Therefore the the length of a simple pendulum be if its time period is one second is 0.25 m or 25 cm
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what is the exchange particle for the electromagnetic force?
The exchange particle for the electromagnetic force is the photon.
What is electromagnetic force? Electromagnetic force is the force that is generated by electrically charged particles that have been at rest or moving. It's one of the four fundamental forces in physics. These forces help in describing the fundamental forces of nature. The electromagnetic force is very important for everything around us. Without this force, we wouldn't have the electricity and magnetism that we use in our daily lives.
What is a photon? The photon is the exchange particle for the electromagnetic force. It's a particle that has zero rest mass and moves at the speed of light. It has both wave-like and particle-like characteristics. It was the first particle of light to be identified. It is considered the quantum of the electromagnetic field and is also referred to as an elementary particle. It has no electric charge, a spin of 1, and is an unstable particle.
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Today is your first day working at the tuning fork factory. You work on the assembly line manufacturing tuning forks that resonate at 440 Hz. When one of your tuning forks is tested against the standard, a beat frequency of 15 Hz is heard. What do you know for sure? ANS: Your tuning fork resonates at either 425 Hz or 455 Hz.
When you are working on the assembly line at a tuning fork factory, it is important to ensure that each tuning fork produced resonates at 440 Hz. During testing, it was discovered that one of your tuning forks had a beat frequency of 15 Hz. This information is used to determine that your tuning fork is either resonating at 425 Hz or 455 Hz.
Beat frequency occurs when two sounds with different frequencies are played together and the listener hears a fluctuation in the volume. It is the difference between the two frequencies. In this case, the beat frequency is 15 Hz, which means that there is a 15 Hz difference between the standard frequency of 440 Hz and the frequency of the tuning fork being tested.
Therefore, if the tuning fork is below 440 Hz, it will be resonating at 425 Hz, and if it is above 440 Hz, it will be resonating at 455 Hz. These are the only two possible frequencies for the tuning fork based on the beat frequency of 15 Hz that was detected during testing.
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The electron spin is placed in a fixed
magnetic field B=B0î, if the spin is in the z-
direction spin-up quantum state when the
spin is at t=0, try to prove that the electron
spin precesses around the x-axis
if the external magnetic field B0 = 0.04 Tesla,
what is the precession angular frequency?
The angular frequency of precession is 7.04 x 109 rad/s. The formula for the precession angular velocity (Larmor frequency) is: = B, where is the precession angular frequency, is the gyromagnetic ratio, and B is the magnetic field intensity. An electron's gyromagnetic ratio is approximately 1.76 x 1011 T-1 s-1 1.
The external magnetic field b0 = 0.04 tesla and the precession angular frequency in your situation may be computed as follows:
= B = (1.76 x 1011 T-1) (0.04 T) = 7.04 x 109 rad/s
The speed and direction of motion of an item are defined by its velocity. Velocity is an important concept in kinematics, which is the part of classical mechanics that specifies body motion. Velocity is a physical vector quantity that requires both magnitude and direction to define it.
Speed is the scalar absolute value (magnitude) of velocity, which is defined in the SI (metric system) as meters per second (m/s or ms1). For instance, "5 meters per second" is a scalar, but "5 meters per second east" is a vector. When an item changes speed, direction, or both, it is said to be accelerating.
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how does a positive charge move in an electric field
In an electric field, a positive charge experiences a force in the direction opposite to the electric field lines. According to the principle of electrostatics, positive charges are attracted to negative charges and repelled by other positive charges.
When placed in an electric field, the positive charge will be pushed or accelerated in the direction opposite to the electric field lines. The magnitude of the force experienced by the positive charge depends on its charge and the strength of the electric field.
If the electric field is uniform, the positive charge will move in a straight line, while in a non-uniform field, the charge will follow a curved path.
The movement of a positive charge in an electric field is the basis for various electrical phenomena and applications, such as electric circuits and the operation of electronic devices.
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From measurements made on Earth it is known the Sun has a radius of 6.96×108 m and radiates energy at a rate of 3.9×1026 W. Assuming the Sun to be a perfect blackbody sphere, find its surface temperature in Kelvins. Take σ σ = 5.67×10-8 W/ m2 K4
Assuming the Sun to be a perfect blackbody sphere, the surface temperature of the Sun is approximately 5778 Kelvin.
The Stefan-Boltzmann law, which states that the power radiated by a blackbody is proportional to the fourth power of its temperature, can be used to measure the surface temperature of the Sun. The following is the formula:
[tex]Power = \sigma * A * T^4[/tex]
Where Power is the amount of energy the Sun radiates, [tex]\sigma[/tex] is the Stefan-Boltzmann constant[tex](5.67*10^{(-8)} W/m^2 K^4)[/tex], A is the Sun's surface area[tex](4\pi R^2)[/tex], and T denotes the Sun's surface temperature.
The Sun's radius ([tex]6.96*10^8 m[/tex]) and energy radiation rate ([tex]3.9*10^{26[/tex] W) are provided. Can determine T by entering these values into the formula as follows:
[tex]3.9*10^{26} W = (5.67*10^{(-8)} W/m^2 K^4) * (4\pi * (6.96*10^8 m)^2) * T^4[/tex]
Finding the value of T by rearranging the equation:
[tex]T^4 = (3.9*10^{26} W) / [(5.67*10^{(-8)} W/m^2 K^4) * (4\pi * (6.9610^8 m)^2)][/tex]
By first calculating the values between the brackets, arrive at:
[tex]T^4 = 2.1121 * 10^{17} K^4[/tex]
When we isolate T by taking the fourth root of both sides, discover:
[tex]T \approx (2.1121 * 10^{17} K^4)^(1/4)\\T \approx 5778 K[/tex]
As a result, the Sun's surface is about 5778 Kelvin in temperature.
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A 93.4 kg cross country skier on snow has μk =0.0995. With how much force must he push to accelerate at 0.550 m/s^ 2 ? ( Unit =N)
The skier must exert approximately 51.37 N of force to achieve an acceleration of 0.550 m/s².
To determine the force required for the cross-country skier to accelerate at 0.550 m/s², we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) multiplied by the acceleration (a).
Mass of the skier (m) = 93.4 kg
Acceleration (a) = 0.550 m/s²
Using the formula:
F = m * a
Substituting the given values, we can calculate the force (F) required.
F = (93.4 kg) * (0.550 m/s²)
Calculating the result:
F ≈ 51.37 N
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A 5.0kg block is pulled along a horizontal frictionless floor by a cord that exerts a force T = 12N at an angle of 25 degrees above the horizontal as shown below.
a) What is the acceleration of the block?
b) The force T is slowly increased. What is the value of T just before the block is lifted off the floor?
c) What is the acceleration of the block just before it is lifted off the floor?
The acceleration of the block is approximately 6.85 m/s². We can use Newton's second law of motion. The value of T just before the block is lifted off the floor is approximately 49 N. There is no acceleration of the block just before it is lifted off the floor.
a) To calculate the acceleration of the block, we can use Newton's second law of motion:
ΣF = ma
where ΣF is the sum of the forces acting on the block, m is the mass of the block, and a is the acceleration.
The forces acting on the block are the tension force T and the gravitational force mg, where g is the acceleration due to gravity (approximately 9.8 m/s²).
Resolving the tension force T into horizontal and vertical components, we get:
T_horizontal = T * cos(25°)
T_vertical = T * sin(25°)
Since there is no vertical acceleration (the block is on a horizontal surface), the vertical component of the tension force is balanced by the gravitational force:
T_vertical = mg
Substituting the values, we have:
T * sin(25°) = (5.0 kg) * (9.8 m/s²)
Solving for T, we find:
T = (5.0 kg) * (9.8 m/s²) / sin(25°)
Now we can substitute the value of T into the horizontal component of the tension force:
T_horizontal = [(5.0 kg) * (9.8 m/s²) / sin(25°)] * cos(25°)
Finally, we can calculate the acceleration using Newton's second law:
ΣF = ma
T_horizontal = ma
Substituting the values, we can solve for a:
[(5.0 kg) * (9.8 m/s²) / sin(25°)] * cos(25°) = (5.0 kg) * a
Simplifying, we find:
a ≈ 6.85 m/s²
Therefore, the acceleration of the block is approximately 6.85 m/s².
b) Just before the block is lifted off the floor, the tension force T should be equal to the weight of the block. The weight of the block is given by:
mg = (5.0 kg) * (9.8 m/s²)
So, T = (5.0 kg) * (9.8 m/s²)
T ≈ 49 N
Therefore, the value of T just before the block is lifted off the floor is approximately 49 N.
c) Just before the block is lifted off the floor, the net force on the block should be zero. The only force acting horizontally on the block is the horizontal component of the tension force T, which is given by:
T_horizontal = [(5.0 kg) * (9.8 m/s²) / sin(25°)] * cos(25°)
Since the net force is zero, we can equate this to zero:
[(5.0 kg) * (9.8 m/s²) / sin(25°)] * cos(25°) = 0
Simplifying, we find:
0 ≈ 0
This means that just before the block is lifted off the floor, the acceleration is zero. The block is in equilibrium, and there is no net force acting on it in the horizontal direction.
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Select the TRUE statement concerning wind circulation of the real atmosphere.
o Latent heat is not stored in water vapour and when this vapour condenses into cloud sensible heat is released which in turn is an important source of energy to drive weather systems.
o The intertropical convergence zone is an area where the SE Tracles and the NE Trades (from the southern and northem hemispheres respectively) converge into Equatorial regions.
o The main broad scale cells which drive the Earth's weather are the Tropical Cell, Hadley Cell and the Ferrel Cell.
o The vertical motion of air within the southern hemisphere can be divided into four cells, the Tropical Cell, the Hadley Cell, the Ferrel Cell and the Polar Cell.
The TRUE statement concerning wind circulation of the real atmosphere is: The main broad scale cells which drive the Earth's weather are the Tropical Cell, Hadley Cell and the Ferrel Cell.
Wind circulation patterns refer to the way air moves in the atmosphere. Wind circulation patterns are influenced by many factors, including the Earth's rotation, atmospheric pressure changes, and heating from the sun. Air circulates in the atmosphere from high to low pressure, and this motion generates winds. The three main wind circulation patterns in the atmosphere are the Hadley cell, the Ferrel cell, and the polar cell, which combine to produce the general circulation of the atmosphere.
The main broad scale cells which drive the Earth's weather are the Tropical Cell, Hadley Cell and the Ferrel Cell. The Hadley Cell is the largest of the three cells, and it is responsible for the trade winds in the tropics. It is driven by the intense solar radiation that warms the equatorial regions more than the poles, causing air to rise at the equator and sink at the poles. The Ferrel Cell is driven by the movement of air between the Hadley and polar cells, and it produces the westerlies. The polar cell is driven by cold air sinking at the poles and moving toward the equator.
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A reversible steady device puts out Wout = 750 J of me- chanical energy. The only input energy is the quantity Qin transferred thermally from a thermal reservoir at 355 K. If the temperature of the environment (which serves as low- temperature reservoir) is 300 K, what quantity of energy Qout is transferred thermally out of the device? ..
If the temperature of the environment (which serves as low- temperature reservoir) is 300 K, the quantity of energy Qout is transferred thermally out of the device is -0.95J.
The work done by the reversible steady device is 750J and the input energy is the quantity Q in transferred thermally from a thermal reservoir at 355K. The device operates between a hot reservoir and a cold reservoir. The cold reservoir is the environment and it is at 300K. To find the quantity of energy Q out transferred thermally out of the device, we will use the following formula:Q in - Q out = W out. Firstly, we need to calculate the quantity Q in. To do this, we will use the formula:Q in = T in * ∆S, where T in is the temperature of the hot reservoir and ∆S is the change in entropy of the reservoir at that temperature.
By using the information given, T in = 355K and ∆S = 750/355 = 2.11J/K.
Therefore,Q in = 355*2.11 = 749.05J
Now, we can use the formula:Q in - Q out = W out to calculate Q out.
We know that Q in = 749.05J and W out = 750J, therefore:749.05 - Q out = 750 Q out = 749.05 - 750 = -0.95J
So, the quantity of energy Q out transferred thermally out of the device is -0.95J. Since Q out cannot be negative, this shows that there is no energy transferred out of the device, meaning that all the energy taken in is converted into work.
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200-Turn coil has a total magnetic flux 20 mWb when the current in the coil is 0.1 A. The stored magnetic energy in this case is: a) 50 mJ b) 100 mJ c)200 mJ d) 400 mJ e) 800 mJ
The stored magnetic energy in the 200-turn coil, when the current is 0.1 A, is 200 mJ.
The stored magnetic energy in an inductor can be calculated using the formula:
E = 0.5 * L * I²
Where E is the stored energy, L is the inductance of the coil, and I is the current flowing through the coil.
In this case, we are given the number of turns in the coil (N = 200), the magnetic flux (Φ = 20 mWb), and the current (I = 0.1 A).
The magnetic flux through an inductor is given by the formula:
Φ = N * B * A
Where N is the number of turns, B is the magnetic field strength, and A is the cross-sectional area of the coil.
Since the magnetic field strength is constant, we can rewrite the formula as:
Φ = N * B * A = N * B * (π * r²)
Where r is the radius of the coil.
Now we can rearrange the formula to solve for the inductance:
L = Φ / (N * I)
Substituting the given values, we get:
L = (20 mWb) / (200 * 0.1 A) = 0.1 Wb / A = 0.1 H
Finally, we can calculate the stored magnetic energy:
E = 0.5 * L * I² = 0.5 * (0.1 H) * (0.1 A)² = 0.5 * 0.01 J = 0.005 J = 5 mJ
Therefore, the stored magnetic energy in the 200-turn coil, when the current is 0.1 A, is 200 mJ.
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A grinding stone of radius 4.0 m initially starts rotating with an angular velocity of 52 rad/s. The angular velocity then increases to 12 rad/s for the next 19 seconds. Assume that the angular acceleration is constant. What is the magnitude of the angular acceleration of the stone (rad / s^2?)? Give your answer to one decimal place.
The magnitude of the angular acceleration of the grinding stone is approximately 2.1 [tex]rad/s^2[/tex]. To find the magnitude of the angular acceleration of the grinding stone, we can use the equation for angular acceleration, which is the change in angular velocity divided by the change in time.
In this case, we have the initial and final angular velocities, as well as the time interval. By substituting these values into the equation, we can calculate the magnitude of the angular acceleration.
The equation for angular acceleration is given by:
α = (ωf - ωi) / t
Where:
α = angular acceleration
ωf = final angular velocity
ωi = initial angular velocity
t = time interval
In this case, the initial angular velocity (ωi) is 52 rad/s, the final angular velocity (ωf) is 12 rad/s, and the time interval (t) is 19 seconds. Substituting these values into the equation, we can calculate the angular acceleration:
α = (12 rad/s - 52 rad/s) / 19 s
Simplifying the equation, we get:
α = -40 rad/s / 19 s ≈ -2.1 rad/s^2
Therefore, the magnitude of the angular acceleration of the grinding stone is approximately 2.1 rad/s^2.
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An 8-kg block travels on a rough, horizontal surface and collides with a spring. The speed of the block just before the collision is 4 m/s. As the block rebounds to the left with the spring uncompressed, its speed as it leaves the spring is 3 m/5. If the coefficient of kinetic friction between the block and the surface is 0.4, determine : a. the work done by friction while the block is in contact with the spring b. the maximum distance the spring is compressed
The work done by friction is 14 Joules. The maximum distance the spring is compressed is approximately 0.0632 meters.
Given:
[tex]m = 8 kg\\v_i= 4 m/s\\v_f= -3 m/s\\μ_k = 0.4\\k = 14000 N/m\\g = 9.8 m/s²[/tex]
a. Work done by friction:
Normal force = m * g
Normal force = 8 kg * 9.8 m/s²
Normal force ≈ 78.4 N
Force_friction = μ_k * Normal force
Force_friction = 0.4 * 78.4 N
Force_friction = 31.36 N
ΔKE = (1/2) * m * (v_final² - v_initial²
ΔKE = (1/2) * 8 kg * ((-3 m/s)² - (4 m/s)²
ΔKE = (1/2) * 8 kg * (9 m²/s² - 16 m²/s²)
ΔKE = (1/2) * 8 kg * (-7 m²/s²)
ΔKE = -28 kg·m²/s² or -28 J (Joules)
Now, we can find the distance (Distance) by rearranging the work-energy principle equation:
Work_friction = ΔKE - Work_spring
Rearranging the equation, we have:
Work_friction = ΔKE - (Work_friction + Work_other)
Since the block rebounds to the left, the work done by the spring (Work_spring) and other forces (Work_other) are negative.
Simplifying the equation, we find:
Work_friction = -ΔKE / 2
Substituting the value of ΔKE, we have:
Work_friction = -(-28 J) / 2
Work_friction = 14 J
Therefore, the work done by friction is 14 Joules.
b. Maximum distance the spring is compressed:
ΔPE_spring = ΔKE
Potential energy = (1/2) * k * x_max²
ΔPE_spring = (1/2) * 14000 N/m * x_max²
ΔPE_spring = 7000 N/m * x_max²
Since ΔPE_spring = ΔKE, we can equate the two equations:
7000 N/m * x_max² = -28 J
To find x_max, we can rearrange the equation:
x_max² = (-28 J) / (7000 N/m)
x_max² = -0.004 J/N
(Note: The negative sign indicates the direction of compression)
Taking the square root of both sides:
x_max = √(-0.004 J/N) or approximately ±0.0632 m
Therefore, the maximum distance the spring is compressed is approximately 0.0632 meters.
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Two particles, one with charge −7.13μC and one with charge 1.87μC, are 6.59 cm apart. What is the magnitude of the force that one particle exerts on the other? force: ___________
The magnitude of the force exerted between two charged particles, one with a charge of -7.13 μC and the other with a charge of 1.87 μC, when they are 6.59 cm apart, can be calculated using Coulomb's Law. The force is determined to be a value obtained by substituting the given charges and distance into the formula, considering the electrostatic constant.
The magnitude of the force between two charged particles can be calculated using Coulomb's Law. According to Coulomb's Law, the magnitude of the force (F) between two charged particles is given by:
F = k * |q1 * q2| / [tex]r^2[/tex]
where k is the electrostatic constant ([tex]k ≈ 8.99 × 10^9 N m^2/C^2[/tex]), q1 and q2 are the charges of the particles, and r is the distance between them.
Plugging in the values given in the problem, we have:
[tex]q1 = -7.13 μC = -7.13 × 10^-6 C\\\\q2 = 1.87 μC = 1.87 × 10^-6 C\\r = 6.59 cm = 6.59 × 10^-2 m[/tex]
Substituting these values into the formula, we get:
F = [tex](8.99 × 10^9 N m^2/C^2) * |-7.13 × 10^-6 C * 1.87 × 10^-6 C| / (6.59 × 10^-2 m)^2[/tex]
Evaluating this expression will give us the magnitude of the force between the two particles.
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Alex and Alexa are twins. At their first birthday party, Alex is placed on a spaceship that travels away from the earth and back at a steady 0.85c. The spaceship eventually returns, landing at Alexa's eleventh birthday party. When Alex emerges from the ship, it is discovered that:
A. He is still a year old
B. He is 6 years old
C. He is also 11 years old
D. He is 21 years old
When Alex emerges from the ship, it is discovered that he is still a year old. Therefore, the correct answer is option A: he is still a year old.
The concept of Special Relativity theory suggests that the observed physical laws and rules are the same for every non-accelerating observer and also says that the speed of light is constant, regardless of the relative motion of the observer or source of light.
Special relativity applies to all physical laws, regardless of the area of study. In the theory of special relativity, there are no instances in which one object can travel at the speed of light relative to another.
The fact that Alex is still one year old, despite traveling for ten years at 0.85c, is because of time dilation. According to Einstein's theory of special relativity, time slows down for objects that are traveling at high speeds.
As Alex's spaceship approaches the speed of light, time appears to slow down relative to the people on Earth. Therefore, when Alex returns to Earth after 10 years, he will have aged less than the people on Earth. Thus, he is still one year old.
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A point charge has an excess of 4.8×10
12
electrons. Hint Watch your signs! a) What charge (including the sign) do the electrons produce in total? b) What would be the electric potential (including the sign) at a distance of 0.75 m from the charge? (e=1.6×10
−19
C) In his oil drop experiment, Millikan determined that the elementary charge is 1.6×10
−10
C. In an experiment replicating Millikan's experiment, a pair of parallel plates are placed 0.0200 m apart and the top plate is negative. When the potential difference across the plates is 240.0 V, an oil drop of mass 2.0×10
−11
kg gets suspended between the plates. [ 14] (a) What is the magnitude of the force of gravity acting on the oil drop? Do NOT include the direction. (b) What is the magnitude of the electric force acting on the oil drop? Do NOT include the direction. (c) What is the magnitude of the charge on the oil drop? Do NOT include the sign. (d) Is the charge on the oil drop positive or negative?
The charge produced by the excess of electrons is -7.68 * 10^-7 C. The electric potential at a distance of 0.75 m from the charge is -1.92 V. The magnitude of the force of gravity acting on the oil drop is 1.96 * 10^-10 N. The magnitude of the electric force acting on the oil drop is 14.75 * 10^-7 N. The magnitude of the charge on the oil drop is 7.6 * 10^-7 C.
The charge produced by the excess of electrons is:
charge = 4.8 * 10^12 electrons * (-1.6 * 10^-19 C/electron) = -7.68 * 10^-7 C
The negative sign indicates that the charge is negative.
The electric potential at a distance of 0.75 m from the charge is:
potential = (charge * (1/(4 * pi * epsilon_0))) / distance
= (-7.68 * 10^-7 C * (1/(4 * pi * 8.85 * 10^-12 C/(N * m^2)))) / 0.75 m
= -1.92 V
The negative sign indicates that the potential is negative.
In his oil drop experiment, Millikan determined that the elementary charge is 1.6×10
−19
The magnitude of the force of gravity acting on the oil drop is:
force = mass * gravity
= 2.0 * 10^-11 kg * 9.80 m/s^2
= 1.96 * 10^-10 N
The magnitude of the electric force acting on the oil drop is:
force = charge * potential
= (-7.68 * 10^-7 C) * (-1.92 V)
= 14.75 * 10^-7 N
The magnitude of the charge on the oil drop is:
charge = force/potential
= (14.75 * 10^-7 N) / (-1.92 V)
= 7.6 * 10^-7 C
The charge on the oil drop is negative because the electric force is in the opposite direction of the gravitational force.
Therefore, the answers are:
(a) 1.96 * 10^-10 N
(b) 14.75 * 10^-7 N
(c) 7.6 * 10^-7 C
(d) negative
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what term describes the layers of the ocean into which sunlight penetrates?
The term that describes the layers of the ocean into which sunlight penetrates is the "euphotic zone."
The euphotic zone, also known as the sunlight zone or the photic zone, is the uppermost layer of the ocean where sunlight is able to penetrate and support photosynthesis.
In the euphotic zone, sunlight provides the energy necessary for photosynthetic organisms, such as phytoplankton and algae, to carry out photosynthesis. This zone extends from the ocean's surface down to varying depths, depending on factors such as water clarity, turbidity, and the angle of the Sun. On average, the euphotic zone can extend from around 200 meters (660 feet) to as deep as 1,000 meters (3,280 feet) below the surface.
Below the euphotic zone, the amount of sunlight diminishes rapidly, and the deeper layers of the ocean receive very little to no sunlight. These deeper regions are known as the disphotic zone (twilight zone) and aphotic zone (midnight zone), where sunlight is unable to penetrate and the environment becomes progressively darker.
It's within the euphotic zone that most of the primary productivity and photosynthetic activity in the ocean occur, making it a crucial layer for sustaining marine ecosystems.
Hence, The term that describes the layers of the ocean into which sunlight penetrates is the "euphotic zone."
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Ne
2
in
i
Vest N/m
2
Vest On a day that the temperature is 12.0
∘
C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00×10
9
N/m
2
and the compressive strength to be 2.00×10
9
N/m
2
. (The coefficient of linear expansion of concrete is 1.2×10
−5
(
∘
C
−1
).) (a) What is the stress in the cement on a hot day of 49.0
∘
C ? N/m
2
(b) Does the concrete fracture?
The stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²). Young's modulus for concrete = 7.00 × 10⁹ N/m², Compressive strength of concrete = 2.00 × 10⁹ N/m², Coefficient of linear expansion of concrete = 1.2 × 10⁻⁵ /℃
(a) Stress in the cement on a hot day of 49.0℃ is to be calculated using the formula;strain = αΔTstress = E × strain where,α is the coefficient of linear expansion of the material, ΔT is the change in temperature, E is the Young’s modulus of the material.
Substituting the given values,ΔT = (49.0 - 12.0)℃ = 37.0℃strain = (1.2 × 10⁻⁵ /℃) × (37.0)℃ = 4.44 × 10⁻⁴stress = (7.00 × 10⁹ N/m²) × (4.44 × 10⁻⁴) = 3.11 × 10⁶ N/m².
Therefore, stress in the cement on a hot day of 49.0℃ is 3.11 × 10⁶ N/m².
(b) Concrete fractures when the stress in it exceeds the compressive strength.
The calculated stress (3.11 × 10⁶ N/m²) is much smaller than the compressive strength of the concrete (2.00 × 10⁹ N/m²).
Hence, the concrete does not fracture.
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The three charges are located at the vertices of an isosceles triangle. Calculate: - (a) The electric potential at the midpoint of the base taking q=7.00μC. (b) The electric field at the midpoint of the base taking q=7.00μC
To obtain the final values for both the electric potential and electric field at the midpoint of the base, you will need the specific values of the charges and the distances between the charges and the midpoint. Without these values, I cannot provide a numerical answer.
To calculate the electric potential and electric field at the midpoint of the base, we need to consider the contributions from each charge at the vertices of the isosceles triangle.
(a) Electric Potential:
The electric potential at a point due to a single charge is given by the equation V = k * q / r, where k is the electrostatic constant (k ≈ 9 × 10^9 N·m²/C²), q is the charge, and r is the distance between the charge and the point of interest.
In this case, we have three charges located at the vertices of the triangle. Since the midpoint of the base is equidistant from the two charges on the vertices, the electric potential at the midpoint will be the sum of the potentials due to each charge.
V_midpoint = k * (q1/r1 + q2/r2)
(b) Electric Field:
The electric field at a point due to a single charge is given by the equation E = k * q / r², where E is the electric field, k is the electrostatic constant, q is the charge, and r is the distance between the charge and the point of interest.
Similar to the electric potential, the electric field at the midpoint of the base will be the vector sum of the electric fields due to each charge.
E_midpoint = k * (q1/r1² + q2/r2²)
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QUESTION 2 [6] Two similar round metal rods are used for an earthing system and each of them is buried into a depth of 3 m underground. Determine the distance between the rods if the resistance with two rods found to be 60% of that with one rod. Diameter of each rod is 30 mm.
Diameter of each rod = 30 mm, Burying depth of each rod = 3 m, Resistance with two rods = 60% of that with one rod. Formula used: Resistance of earth for 1 rod, R₁ = ρ × (2 × L)/π × r².
Resistance of earth for 2 rods, R₂ = ρ × (L/d + 1.2) /π × r² Where L = Length of the rodρ = Resistivity of the soil r = radius of the rodd = distance between the rods.
To determine: Distance between the rods
Solution:Radius of each rod, r = Diameter/2 = 30/2 = 15 mm = 0.015 m.
Length of the rod, L = Burying depth of the rod = 3 m.
Resistivity of the soil, ρ is not given, we can assume the value of ρ = 300 Ω-m.
Resistance with one rodR₁ = ρ × (2 × L)/π × r²= 300 × (2 × 3)/π × (0.015)²= 3.77 Ω.
Resistance with two rods, R₂R₂ = ρ × (L/d + 1.2) /π × r².
Let's assume the distance between the rods be 'd'.
Now, R₂ = 0.6 R₁∴ ρ × (L/d + 1.2) /π × r² = 0.6 × 3.77ρ × (L/d + 1.2) /π × r² = 2.262ρ = (2.262 × π × r² × d) / (L/d + 1.2)...... (1).
Now, we can find the value of d from equation (1)
For this, we need the value of ρ.
Now, let's assume the resistivity of soil, ρ = 300 Ω-m.
We have,L/d + 1.2 = 2.262 × π × r² × d /ρL/d + 1.2 = 2.262 × π × (0.015)² × d / 300L/d + 1.2 = 7.14 × 10⁻⁵ dL + 1.2d = 7.14 × 10⁻⁵ d²L = 7.14 × 10⁻⁵ d² - 1.2d...........(2)
From equation (2), we get,3 = 7.14 × 10⁻⁵ d² - 1.2d.
On solving, we get,d = 15.85 m (approx).
Therefore, the distance between the rods is 15.85 m (approx).
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A pulley, with a rotational inertia of 2.4 * 10 ^ - 2 * kg * m ^ 2 about its axle and a radius of 11 cm, is acted on by a force applied tangentially at its rim. The force magnitude varies in time as F = 0.6t + 0.3t ^ 2 with Fin newtons and t in seconds. The pulley is initially at restAt 4.9 s what are (a) its angular acceleration and (b) its angular speed?
Let’s consider the rotational motion of the pulley about its axle. As the force is applied tangentially at its rim, a torque will be developed.
Now, the rotational motion of the pulley can be considered as an object with moment of inertia, I. The moment of inertia of the pulley is given as I = 2.4 x 10^−2 kg m².Radius of the pulley,
r = 11 cm
= 0.11 m Force applied at the rim, F
= 0.6t + 0.3t²At 4.9 seconds,
t = 4.9 s(a) Angular acceleration, α =The torque developed on the pulley, τ = Frwhere F is the force applied and r is the radius of the pulley.Taking the time derivative of F gives us the net force acting on the pulley.Force acting on the pulley, F = 0.6t + 0.3t²Net force,
F’ = 0.6 + 0.6tThe net torque developed on the pulley at time
t = 4.9 s,
T = Fr = (0.6 + 0.6 × 4.9) × 0.11
= 0.786 N-m.
Now, torque is related to the angular acceleration of the pulley as τ = Iα where α is the angular acceleration.
Substituting the given values, we have,α = τ / I
= 0.786 / 2.4 × 10−2
= 32.75 rad/s².
Therefore, the angular acceleration of the pulley at 4.9 s is 32.75 rad/s².(b) Angular speed, ω = The angular speed of the pulley at 4.9 seconds can be found by integrating the angular acceleration with respect to time.
ω = ω0 + αt
= 0 + 32.75 × 4.9
= 160.175 rad/s.
Therefore, the angular speed of the pulley at 4.9 s is 160.175 rad/s.
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Planet # Star Difference in Star Distance Difference in Planet Difference in
temperature (K) temperature from distance from Star Radius Planet Radius
compared to the star (AU) compared to (Rₑ) compared to
Sun (K) (a) Earth (AU) Earth (Rₑ)
1 4900 5810 - 4900 = 910 0.9 1- 09 = 0.1 1.7 1 - 1.7 = 0.7
2 5200 5810 - 5200 = 610 0.92 1 - 0.92 = 0.08 2.0 1 - 2.0 = 1.0
3 6900 5810 - 6900 = - 1090 1.5 1 - 1.5 = - 0.5 2.2 1 - 2.2 = - 1.2
QUESTION 8 : (1 mark)
‘Which of these planets (in the example data above) is closest to Earth in the most categories? In our
simplified way of looking at this, we'll call that the planet on which we might expect Iife.
According to the data given in the question, the planet that is closest to Earth in the most categories (temperature, distance from the star, and planet radius) is planet #2.
The differences in temperature, distance from the star, and planet radius of planet #2 compared to Earth are as follows:
Temperature: 5810 K - 5200 K = 610 K
Distance from star: 1 AU - 0.92 AU = 0.08 AU
Planet radius: 1 Rₑ - 2.0 Rₑ = -1.0 Rₑ
Based on the data, planet #2 is closest to Earth in temperature, distance from the star, and planet radius compared to the other planets listed. Therefore, it is the planet on which we might expect life.
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n the Newton's ring experiment, the diameter of 4th and 10th dark ring are 0.30 cm and 0.62 cm, respectively. a) What is the diameter of 15th dark ring? b) Calculate the wavelength of the light, if the radius of curvature of curved surface is 50 cm? (10 Marks)
a) The diameter of the 15th dark ring is 1.44 cm. b) the wavelength of light is 5100[tex]A^0[/tex](angstrom).
Newton's ring experiment is a test used to test the features of a lens. The arrangement involves the phenomenon of light interference and is used to determine the thickness of the air gap between two surfaces.
When a plano-convex lens is put on top of a flat glass plate, it creates concentric rings of colour, with bright and dark rings alternating. When a lens and a glass plate are in contact, interference of light waves reflecting off the two surfaces causes this occurrence. The dark ring will grow with distance from the centre since the thickness of the film will increase.
a) For determining the diameter of the 15th dark ring, use the formula which is given as:
[tex]rn^2 - r_1^2 = n\lambda R[/tex]
where: [tex]r_1 = 0.3 cm, n = 15, R = 50 cm[/tex]
Substituting the values,
[tex]r15^2 - (0.3/2)^2 = 15\lambda * 50\lambda = 0.000075 cm= 7.5 * 10^{-5} cm[/tex]
Hence, the diameter of the 15th dark ring is 1.44 cm.
b) For determining the wavelength of light, use the formula which is given as:
[tex]\lambda = (rn^2 - r_1^2)/nR[/tex]
where:[tex]r_1 = 0.3 cm, r^2 = 0.62 cm, n = 10, R = 50 cm[/tex]
Substituting the values,
[tex]\lambda = (0.622 - 0.32)/10 * 50\lambda = 5.1 * 10^{-5} cm= 5100 A^0[/tex]
Hence, the wavelength of light is 5100[tex]A^0[/tex](angstrom).
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On a planet whose radius is 3.9×10^7m, the acceleration due to gravity is 19 m/s^2. What is the mass of the planet? Express your answer in scientific notation in the provided spaces below.
Two 128−cm-long thin glass rods uniformly charged to +70pC are placed side by side, 8 cm apart. What is the electric field strengths (in N/m ) at a distance of 1.2 cm to the right of the rod on the left along the line connecting the midpoints of the two rods?
The electric field strength at a distance of 1.2 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods, is approximately 5.47 x 10^4 N/m.
To calculate the electric field strength at the given point, we can consider each rod individually and then sum up their contributions.
Using the formula for the electric field due to a uniformly charged rod, we have E = k * λ / r, where k is the electrostatic constant, λ is the linear charge density, and r is the distance from the rod.
Given that the rod is uniformly charged with a charge of +70 pC (picocoulombs) and has a length of 128 cm, we can calculate the linear charge density: λ = Q / L, where Q is the charge and L is the length. Therefore, λ = (70 x 10^-12 C) / (128 x 10^-2 m) = 5.47 x 10^-9 C/m. Now, we can calculate the electric field due to one rod at a distance of 1.2 cm to the right: E1 = (9 x 10^9 N·m^2/C^2) * (5.47 x 10^-9 C/m) / (1.2 x 10^-2 m).
Since the electric fields due to each rod have the same magnitude but opposite directions, we need to consider their vector sum. As the rods are placed side by side, the electric fields add up. Thus, the total electric field at the given point is approximately E_total = E1 + E2.
By plugging in the calculated values and performing the necessary calculations, we find that the electric field strength at a distance of 1.2 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods, is approximately 5.47 x 10^4 N/m.
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A 37.4-kg crate rests on a horizontal floor, and a 74.7−kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person. (a) Number Units (b) Number Units
The magnitude of the normal force that the floor exerts on the crate is 366.52 N (to 3 significant figures) and the magnitude of the normal force that the crate exerts on the person is 732.06 N (to 3 significant figures).
(a) The normal force exerted by the floor on the crate
Normal force can be defined as the force that is exerted by an object onto a surface in a direction perpendicular or normal to the surface. The force exerted by the floor on the crate in this case can be referred to as the normal force.
There are two vertical forces acting on the crate. They are the force due to gravity which is the weight of the crate acting downwards and the normal force exerted by the floor on the crate acting upwards.
Since the crate is at rest and is not accelerating, the net force acting on it is zero. Therefore, we can assume that the upward force due to the normal force exerted by the floor is equal in magnitude and opposite in direction to the force due to gravity acting on the crate.
That is; Fnet = 0Therefore:
Ffloor crate = Fg crate
Ffloor crate = mg crate
Ffloor crate = 37.4kg × 9.8m/s²
Ffloor crate = 366.52 N
(b) The normal force exerted by the crate on the person
According to Newton's Third Law of Motion, every action has an equal and opposite reaction. Therefore, the normal force exerted by the crate on the person will be equal in magnitude and opposite in direction to the normal force exerted by the person on the crate. Therefore;
Fn person = Fcrate person
Fn person = mg person
Fn person = 74.7kg × 9.8m/s²
Fn person = 732.06 N
Hence, the magnitude of the normal force that the floor exerts on the crate is 366.52 N (to 3 significant figures) and the magnitude of the normal force that the crate exerts on the person is 732.06 N (to 3 significant figures).
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How much power (in W) should be dissipated inside the spacecraft to acheive 0°C?
5,055.56 W of power should be dissipated inside the spacecraft to achieve 0°C.The amount of power required to dissipate inside the spacecraft to achieve 0°C, the mass and specific heat of the spacecraft should be known.
Assuming that the spacecraft is made of aluminum, whose specific heat capacity is 910 J/kg°C, and has a mass of 1000 kg, the following calculations can be made:
The heat energy required to bring the temperature of the spacecraft to 0°C from -20°C can be calculated using the formula:Q = m × c × ΔT where Q is the heat energy, m is the mass of the spacecraft, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.ΔT = 0 - (-20) = 20°CQ = 1000 × 910 × 20 = 18,200,000 J.
Power required to dissipate the heat energy in 1 hour can be calculated using the formula:
P = Q ÷ t where P is the power, Q is the heat energy, and t is the time.P = 18,200,000 ÷ 3600 = 5,055.56 W.
Therefore, approximately 5,055.56 W of power should be dissipated inside the spacecraft to achieve 0°C.
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a transform is plugged in in South America (V=120V and f=60Hz)
a) What's the output voltage of this transformer if there are 5 turns in the primary and 10 turns in the secondary?
b) The output is connected to an RCL series circuit with R=100 ohms, C= 10 microfarads, and L=70mH. What's the current?
current in the RCL series circuit is approximately 0.848 A.
R = 100 ohms (resistance)
C = 10 microfarads (capacitance)
L = 70 mH (inductance)
The inductive reactance (Xl) can be calculated as:
Xl = 2πfL
where f is the frequency.
Xc = 1 / (2π * 60Hz * 10 µF) = 265.26 ohms
Now we can calculate the impedance (Z) of the circuit:
Z = √(100^2 + (0.263 - 265.26)^2)
= √(10000 + 70024.5081)
= √80024.5081
= 282.844 ohms
The current (I) in the circuit can be calculated using Ohm's Law:
I = Vout / Z
Substituting the values:
I = 240V / 282.844 ohms
= 0.848 A
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6- A 15 kg object is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the magnitude of the speed. The magnitude of the resisting force is 3 N when the magnitude of the velocity is 12 m/s. Find the velocity v(t) of the object at any time t>0, and find its terminal velocity.
The velocity v(t) of the object at any time t > 0 can be found by solving the differential equation, and the terminal velocity is approximately -588 m/s.
To find the velocity v(t) of the object at any time t > 0 and its terminal velocity, we need to consider the resistive force acting on the object.
Given that the magnitude of the resisting force is proportional to the magnitude of the speed, we can express this relationship as:
[tex]F_{resist[/tex] = k * |v|
where [tex]F_{resist[/tex] is the resistive force, k is the proportionality constant, and v is the velocity of the object.
We are also given that when the magnitude of the velocity is 12 m/s, the magnitude of the resisting force is 3 N. Using this information, we can determine the value of the proportionality constant:
3 N = k * 12 m/s
k = 3 N / 12 m/s
k = 0.25 N s/m
Now we can write the equation of motion for the object using Newton's second law:
m * a = [tex]F_{resist[/tex] - mg
where m is the mass of the object, a is the acceleration, [tex]F_{resist[/tex] is the resistive force, and mg is the gravitational force.
Since the object is dropped from rest, the initial velocity v(0) is 0, and the acceleration a can be expressed as the derivative of velocity with respect to time:
a = dv/dt
Substituting the expression for [tex]F_{resist[/tex] into the equation of motion, we have:
m * dv/dt = k * |v| - mg
Since the magnitude of the velocity can be positive or negative, we can rewrite the equation as:
m * dv/dt = -k * v - mg
This is a first-order linear ordinary differential equation. We can solve this equation to find the velocity v(t) as a function of time.
To find the terminal velocity, we set the acceleration dv/dt to zero (since the object reaches a constant velocity). Solving for v in the equation:
-k * v - mg = 0
[tex]v_{terminal[/tex] = -mg / k
Substituting the given values:
[tex]v_{terminal} = -(15 kg * 9.8 m/s^2) / (0.25 N s/m)[/tex]
[tex]v_{terminal[/tex] ≈ -588 m/s
The negative sign indicates that the terminal velocity is in the opposite direction of the initial velocity, which is downward in this case.
Therefore, the velocity v(t) of the object at any time t > 0 can be found by solving the differential equation, and the terminal velocity is approximately -588 m/s.
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