Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum
What is Faraday's law?Faraday's law states that a changing magnetic field through an area or equivalently, a changing area with constant field will cause a voltage.
So therefore, Faraday's law is applicable in the mechanism behind electric generators, credit cards, metal detectors, computer hard drives and electronic drum
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
The orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s
What is law of gravitation?The law of gravitation states that the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between the masses. Mathematically;
F = GMm/r²
where
M and m are the mass of ice cube and
Recall that;
s = Gm1/r^2
Also;
F = sm²
Substitute to have;
s = m²/F
For the centripetal acceleration
a = v²/r
Such that;
v²/r = Gm/r²
v² = Gm/r
v = √Gm/r
Substitute the given parameters into the formula to have:
V = √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5
V = 355358.97m/s = 3.56 * 10^6 m/s
Therefore the orbital speed of an ice cube in the rings of saturn is approximately 3.56 * 10^6 m/s
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"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it mean?
Answer: It simply means that a freely falling object would increase its velocity by 9.8 m/s per second
Answer:
Hello!
"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that
That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.
The cricket ball has a mass of 0.16kg and it hits the bat with a speed of 25 m/s. After being in contact with the bat for 0.0013 s, the ball rebounds with a speed of 22 m/s in the direction exactly opposite to its original direction.
A) state the difference between speed and velocity
B) calculate.
I) the change in velocity of the cricket ball
Ii) the average acceleration of the ball whilst it is in contact with tge bat
Iii) the average force exerted on the ball by the bat
change in velocity is 3m/s,the force exterted by bat on ball is 4.8N and the acceleration is 30m/s².
Give some difference between velocity and speed.1) Speed is scalar quantity but velocity is vector quantity.
2) Speed is distance followed by an individual with respect to time and velocity is displacement with respect to time .
1) final velocity- intial velocity
25-22= 3m/s
2) The force is
F= ma
. 0.16×30= 4.8N
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a 30kg crate is pulled up a ramp 15m long and 2m high by a constant force of 100n. the crate starts from resta nd has a velocity of 2m/s when it reaches the ramp. what is the frictional force between the crate and the ramp. use the principal of conservation energy.
The frictional force between the crate and the ramp is determined as 35.2 N.
Energy lost to frictional forceApply the principle of conservation energy to calculate the change in the energy of the crate.
Change in energy of the crate = energy loss to friction
P.Ei - K.Ef = E
mgh - ¹/₂mv² = E
where;
m is mass of the crateh is vertical height traveled by the cratev is the final velocity of the crate(30)(9.8)(2) - (0.5)(30)(2²) = E
528 J = E
Frictional force between the crate and the rampE = Fd
where;
F is the frictional forced is the distance traveled by the crateF = E/d
F = (528)/(15)
F = 35.2 N
Thus, the frictional force between the crate and the ramp is determined as 35.2 N.
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figue 1 shows a piece of elastic being stretched between 2 pieces of wood
When a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed
However, in physics, the type of stress applied when an elastic material is stretched is tensile stress.
Recall:
Stess is defined as force per unit area
Mathematically; Stress = F/A
What is elasticity?Elasticity can be defined as the ability of a deformed elastic material or body to return to its original size and shape when the forces causing the deformation are removed.
So therefore, when a piece of elastic material between two pieces of wood is being stretched beyond its limit, the elastic material or the object does not return to its original length when the force is removed
Complete question:
What happens when a piece of elastic material between two pieces of wood is being stretched beyond its limit?
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An Airbus A380-800 passenger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m^2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? (The average density of the air around the airplane is ρair = 1.21 kg/m^3.)
The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:[tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex] We substitute d for to represent density.
We've done that,[tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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(refer to photos attached. Example of previous question with wrong/correct answers example, and current question needing to be solved)
Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −3.94 µC is placed at this point, what are the magnitude and direction of the force on it?
Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
(a) The electric field strength at a point 1.00 cm to the left of the middle is 2.0 x 10⁷ N/C.
(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first chargeE1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second chargeE2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third chargeE3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)
E = +2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = -94.4 N
Thus, the direction of the force will be towards the left.
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Physics occurs all the time but often goes unnoticed. Here is your chance to reflect on physics in action. Other than the examples used in this lesson, think of a time where you witnessed the conservation of angular momentum. Describe the objects that had angular momentum and how angular momentum was conserved. You may also create an example if you cannot recall one in your personal experience.
Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
What is principle of conservation of angular momentum?The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no external torque acting on the object.
Angular momentum of a system is conserved as long as there is no net external torque acting on the system.
Examples of conservation of angular momentumthe spinning ice skatersomeone spinning in an office chaira child spinning on roller coasterThus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
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a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
The correct answer is 5828.675 J.
Given combined mass 4kg and mass of bullet 150gm=0.150kg.
Total mass= 4+0.150=4.150kg
Velocity=53 m/s
Kinetic energy = [tex]\frac{1}{2} *m*v^{2}[/tex] =0.5*4.150*[tex]53^{2}[/tex] =5828.675 J
Kinetic energyKinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by being subjected to a net force, it accelerates and consequently obtains kinetic energy. A moving object or particle's kinetic energy, which depends on both mass and speed, is one of its characteristics. Any combination of motions, including translation (or travel along a path from one location to another), rotation about an axis, and vibration, may be used as the type of motion.
A body's translational kinetic energy is equal to [tex]\frac{1}{2} *m*v^{2}[/tex] , or one-half of the product of its mass, m, and square of its velocity, v.
a trolley and a sandbag have a combined mass of 4 kg. a bullet with a mass of 150 gram is fired towards the trolley and is lodged in the sand bag immediately after the Collision, on the trolley and sandbag in combination with the bullet, move backwards at 53 metre per second calculate the kinetic energy of the trolley , with the sandbag and bullet ,directly after the Collision
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The resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20 a, what is the voltage across the wire?
The voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
What does the Resistance of a wire depend on ?The resistance of a wire is the opposition to the flow of current. It depends on the following;
TemperatureLength of the wireCross sectional areaResistivity of the wireGiven that the resistivity of a 1.0 m long wire is 1.72 × 10-8 ωm and its cross sectional area is 2.0 × 10-6 m2. if the wire carries a current of 0.20A
The given parameters are;
Resistivity ρ = 1.72 × 10-8 ωm Length L = 1.0 mCross sectional area A = 2.0 × [tex]10^{-6}[/tex] m²Current I = 0.2 AResistance R = ?Voltage V = ?The formula to use to get R will be
R = ρL / A
Substitute all the necessary parameters into the formula
R = 1.72 x [tex]10^{-8}[/tex] x 1 / 2 x [tex]10^{-6}[/tex]
R = 8.6 x [tex]10^{-3}[/tex] Ω
From Ohm's law, V = IR
Substitute all the necessary parameters into the formula
V = 0.2 x 8.6 × [tex]10^{-3}[/tex]
V = 1.72 x [tex]10^{-3}[/tex] V
Therefore, the voltage across the wire is 1.72 x [tex]10^{-3}[/tex] V
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ASAP NEED HELP !! :(
Why are temperatures more moderate around the fall and spring equinoxes?
C The angle at which Earth's axis tilts changes.
Neither end of Earth's axis is tilted toward the Sun.
The north end of Earth's axis is tilted toward the Sun.
C
The Earth briefly wobbles on its axis.
Answer:C is the answer
Explanation:
It is the most reasonable and the answer that makes most sense
Answer:
The angle at which Earth's axis tilts changes.
*please refer to photo*
Use meters and seconds for all calculations. Show the appropriate units for measured and calculated values
Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.
What is average velocity?An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.
What is an average?An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.
How to calculate the average velocity?Mathematically, the average velocity for this data set would be calculated by using this formula:
Average = [F(v)]/n
Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5
Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:
v₁ = 100 m/sv₂ = 150 m/sv₃ = 200 m/sv₄ = 250 m/sv₅ = 300 m/sSubstituting the parameters into the formula, we have:
Vavg = [300 + 450 + 500 + 250 + 300)/5
Vavg = 1800/5
Vavg = 360 m/s.
Next, we would calculate the percent difference by using this formula:
[tex]Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100[/tex]
Percent difference = [360 - 343]/360 × 100
Percent difference = 17/360 × 100
Percent difference = 0.0472 × 100
Percent difference = 4.72%.
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If the velocity of an object is -8 m/s and its momentum is -32 kgm/s, what is its mass?
The revolution of the earth around the sun demonstrate what motion?
Answer:
Anticlockwise directions
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You are designing a delivery ramp for crates containing exercise equipment. The 1390 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0∘. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp.
The maximum force constant of the spring Kmax is 2337.9 N/m.
What is force constant of a spring?The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Force = K * xWhere;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
The required orbital speed of the ice cube is 355,358m/s
What is gravitational law?The force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. This can be expressed mathematically as;
Fr = GMm/r²
The distance is calculated as;
s = Gm/r²
Solving both equation, we will have:
v²/r = Gm/r²
v² = Gm/r
Take the square root of both sides
v = √Gm/r
Solve the required orbital speed
V = √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5
V = 355358.97m/s
Hence the required orbital speed of the ice cube is 355,358m/s
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A visitor at a Michigan lighthouse is trying to measure the height of the lighthouse. She has a big spool of string, but she doesn't have a measuring tape. She has a wrist watch with a timer function on it. She attaches a weight to the end of the string and she makes a simple pendulum. She hangs the pendulum down from the top of the spiral staircase and she measures the period of the pendulum. The period of the pendulum is 8.48 s. What is the height of the light house?
The height of the light house is 17.86m.
To find the answer, we have to know about the simple pendulum.
How to find the height of the light house?We have the expression for time period of the pendulum as,[tex]T=2\pi \sqrt{\frac{l}{g} } \\[/tex]
where, g is acceleration due to gravity, and l is the length of the pendulum.
For the pendulum with time period T kept at some height under the influence of gravity, then the height will be equal to,[tex]h=(\frac{T}{2\pi }) ^2g=(\frac{8.48}{2*3.14}) ^2*9.8=17.86m[/tex]
Thus, we can conclude that, the height of the light house is 17.86m.
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iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weight of iron ball in air?
Answer:
453 gm
Explanation:
Immersed objects are buoyed up by force equal to mass of displaced liquid
400 + 53 = 453 gm in air
How much energy is needed to move an electron in a hydrogen atom from the ground state (n = 1) to n = 3?
The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.
How to compute the value?The following can be deduced:
Energy of electron in hydrogen atom is
En = -13.6 /n2 eV
where n is principal quantum number of orbit.
Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV
Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV
Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1
ΔE = -1.51 - ( 13.6) = 12.09 eV
Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.
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Complete question:
How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.
A jet of water squirts out horizontally from a hole near the bottom of the tank, as seen in the figure.
If the hole has a diameter of 3.13 mm, what is the height, h, of the water level in the tank? Assume that x = 1.33 m and y = 1.72 m.
The height of the water level in the water tank is 0.011 m.
What is the time taken by a fluid particle to reach the ground?From Newtown's equation of motion, y=U×t+(1/2)gt²As, initial velocity (U) is zero, t= 2y/gt = (2× 1.72)/9.8 = 0.35 s
What's the velocity along x direction?Velocity along x direction= x × t
= 1.33 × 0.35
= 0.46 m/s
What's the height of water level in the tank?As per Bernoulli's theorem, 1/2 × ρ × v² = ρ × g× h
=> h = v²/2g = 0.46²/(2×9.8)
= 0.011 m
Thus, we can conclude that the height of water level is 0.011m.
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The velocity of a body is given by the equation v= a + bx, where 'x' is displacement. The unit of b is .......
Answer:
s^ -1 ( or 1/sec)
Explanation:
Velocity is given in units of displacement / sec
like feet /sec or m/sec
so b would have units of s^-1
(or perhaps a more general term would be time^-1)
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.
The change in the speed of the space capsule will be -0.189 m/s.
The average force exerted by each on the other will be 567 N.
The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
Given:Mass of the astronaut, [tex]m_a[/tex] = 126 kg
Speed he acquires, [tex]v_{a}[/tex] = 2.70 m/s
Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg
The initial momentum of the astronaut-capsule system is zero due to rest.
[tex]P_f = m_av_a + m_cv_c[/tex]
[tex]P_I[/tex] = 0
[tex]m_av_a + m_cv_c = 0[/tex]
[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]
[tex]= \frac{126* 2.70}{1800}[/tex]
[tex]= - 0.189[/tex] m/s
Therefore,
According, to the impulse-momentum theorem;
FΔt = ΔP
ΔP = m Δv
ΔP = 126×2.70
= 340.2 kgm/sec
t is time interval = 0.600s
F = ΔP/Δt
F = 340.2/0.600
= 567 N
Therefore, the average force exerted by each on the other will be 567 N.
The Kinetic Energy of the astronaut;
K.E = [tex]\frac{1}{2} m v^2[/tex]
[tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]
= 459.27 J
The Kinetic Energy of the capsule;
K.E = [tex]\frac{1}{2} m v^2[/tex]
= [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]
= 32.14 J
Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.
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An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S
Answer:
100rad because it angular velocity
3 A rocket of mass 1200 kg is travelling at 2000 m/s. It fires its engine for 1 minute. The forward thrust provided by the rocket engines is 10 kN (10 000 N).
(i) Use increase in momentum = F x t to calculate the increase in momentum of the rocket.
(ii) Use your answer to a to calculate the increase in velocity of the rocket and its new velocity after firing the engines.
The impulse shared by the object equals the difference in momentum of the object. In equation form,
F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.
How to calculate thrust provided by the rocket engines is 10 kN (10 000 N).?
a)There is this impulse-momentum change equation.
[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]
We consider everything exits happen along a straight line, and gravitation does not participate.
So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.
[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]
New velocity after engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.
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An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 1.9 times the mass of piece B. The energy of 7900 J is released in the explosion.
a)Determine the kinetic energy of piece A after the explosion.
Express your answer to two significant figures and include the appropriate units.
b)Determine the kinetic energy of piece B after the explosion.
Express your answer to two significant figures and include the appropriate units.
Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.
What is the relation between the masses of A and B?Let mass of piece A = MaMass of piece B = Mb
Velocities of pieces A and B are Va and Vb respectively.As per conservation of momentum,Ma×Va = Mb×Vb
Here, Ma=1.9MbSo, 1.9Mb × Va = Mb×Vb
=> 1.9Va = Vb
What are the kinetic energy of piece A and B?Expression of kinetic energy of piece A = 1/2 × Ma × Va²Kinetic energy of piece B = 1/2 × Mb × Vb²Total kinetic energy= 7900J=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900
=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900
=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j
=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule
Kinetic energy of piece B = 7900 - 2724 = 5176 JouleThus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.
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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then
(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s
(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.
c) The speed of the 0.950- kg puck after the collision is 2.82 m/s
d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.
Given:Mass of ice puck, m₁ = 0.350 kg
Mass of another puck, m₂ = 0.950 kg
Velocity of ice puck, v₁ = 5.22 m/s
Velocity of another puck, v₂ = 0 m/s
[tex]v^{'}_1= ?[/tex]
[tex]v^{'}_2= ?[/tex]
[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]
[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]
[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]
Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.
[tex]v^{'}_2 = v^{'} + v_1[/tex]
[tex]= -2.40+5.22[/tex]
[tex]= 2.82 m/s[/tex]
Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.
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Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.3 min to turn through 22 complete revolutions before reaching its final speed.
a)What was its angular acceleration (assumed constant)?
Express your answer using two significant figures.
b)What was its final angular speed in rpm ?
Express your answer using two significant figures
(a) The angular acceleration will be 26.035 rev/[tex]min^{2}[/tex].
(b) The final angular velocity is expected to be 33.846 rev/min.
Given.
t=1.3 min, Θ=22 rev, [tex]ω_{i}[/tex]=0
We know, Θ= [tex]ω_{i}[/tex]t+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]t^{2}[/tex]
22=0+[tex]\frac{1}{2}[/tex][tex]\alpha[/tex][tex]1.3^{2}[/tex]
[tex]\alpha[/tex]=26.035 rev/[tex]min^{2}[/tex]
[tex]ω_{f} =ω_{i}+\alpha t[/tex]=0+26.035*1.3=33.846 rev/min
Angular velocityAn object's rate of change in angular position or orientation over time is depicted by its angular velocity, rotational velocity, or both ( or ), also known as the angular frequency vector (i.e. how quickly an object rotates or revolves relative to a point or axis). The direction of the pseudovector is normal to the instantaneous plane of rotation or angular displacement, and its magnitude denotes the angular speed, or the rate at which the item rotates or revolves. It is customary to use the right-hand rule to specify the direction of angular motion. A general definition of angular velocity is "angle per unit time" (angle replacing distance from linear velocity with time in common). Radians per second is how angles are measured in the SI.
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Is India a rich country?
Explanation:
India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.
Answer:
50505050128128128128
Explanation:
A fighter plane is flying overhead at mach 1.20. What angle does the wave front of the shock wave produced make relative to the plane's direction of motion (in degrees)?
Answer: 56.44°
Explanation:
Given:
Let u represent the current speed of the plane, 1.2 MachConverting to SI Units (m/s):
= (1.2 mach)(340 ms^-1 / 1 Mach)
u = 408 m/s
Speed of sound in air, v = 340 m/sFind:
Angle the wave front of the shock wave relative to the plane's direction of motion, θWe have, sinθ = speed of sound / speed of object
sinθ = v / u
θ = sin^-1 (v / u)
= sin^-1 (340 / 408)
θ = 56.44°
Which of the following statements are true about gravity? Check all that
apply.
A. Gravity exists between two objects that have mass.
B. Gravity exists in the whole universe.
C. Gravity doesn't exist between Earth and the sun.
D. Gravity is a force that pulls two objects together.
E. Gravity exists only on Earth.
SUBMIT
Answer:
Gravity exist between two objects that have mass