Aster's final position from her initial position is 63 m approximately. She will head north west direction to return to her initial position
What is Displacement ?Displacement is the distance travelled in a specific direction. It is a vector quantity.
Given that a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
a) Let P be the Aster's final position from her initial position?
We can use bearing by using Cosine formula to solve this question.
P² = 70² + 82² - 2 × 70 × 82 cos 73
P² = 4900 + 6724 - 11480 cos 73
P² = 11624 - 3356.43
P² = 8267.57
P = √8267.57
P = 90.9 m
P = 90.9 - 28
P = 62.9 m
We can get the angle by using Sine rule
82/ sin Ф = 90.9 / sin 73
sin Ф = 0.8627
Ф = [tex]Sin^{-1}[/tex] (0.8627)
Ф = 59.6°
Ф = 60°
b) She will head north west direction to return to her initial position
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A 29-g rifle bullet traveling 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.5-m-long string, which makes the pendulum swing upward in an arc. (Figure 1)
a) Determine the vertical component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
b)Determine the horizontal component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
What is the velocity of the pendulum just after the collision?As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity
=> 0.029kg × 200 = 3.429 × velocity of pendulum
=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s
What is the amplitude of the pendulum's motion?As per conversation of energy,1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²
Here, k = mg/L = (3.429×9.8)/2.5 = 13.44 N/mSo, 3.429×1.7²= 13.44× amplitude²=> 10 = 13.44× amplitude²
=> Amplitude = √(10/13.44)
= 0.86m
What is the angle made by the pendulum with vertical direction at largest displacement?Angle = tan inverse (amplitude/length of pendulum)
= tan inverse (0.86/2.5)
= 20°
What is the maximum displacement along vertical direction?Maximum displacement along vertical direction= L - L×cos20°
= 2.5 - 2.5×cos20°
= 0.15 m
What's the maximum displacement along horizontal direction?Maximum displacement along horizontal direction= Lsin20°
= 2.5 × sin 20°
= 0.85m
Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.
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how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
2.89watts.
What is meant by sound intensity?The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:
Formula: [tex]I\frac{P}{4\pi r^{2} }[/tex]
I=1.6x10-3 w/m2
r=12m
[tex]I\frac{P}{4\pi r^{2} }[/tex]
[tex]P=I4\pi r^{2}[/tex]
[tex]=(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )[/tex]
The power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m [tex]=2.89watts.[/tex]
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Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com
Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =
(b) What are the x and y components of the electric field
at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =
(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?
magnitude
N/C
direction
° counterclockwise from the +x-axis
(d) What is the magnitude of the force on B3 due to this net electric field?
Given the following data:
Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.How to determine the x and y components of the electric field?First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.
The electric field due to B₁ to B₃ is given by:
E₁ = kq₁/r₁²
E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)
E₁ = 153 × 10³/0.0125
Electric field, E₁ = 12.24 × 10⁶ N/C.
The electric field due to B₂ to B₃ is given by:
E₂ = kq₂/r₂²
E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)
E₂ = 45 × 10³/0.01
Electric field, E₂ = 4.5 × 10⁶ N/C.
Also, the magnitude of the angle formed is given by tan trigonometry:
Tanθ = 5/10
Tanθ = 0.5
θ = tan⁻¹(0.5)
θ = 26.56°.
Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁x = E₁cosθ - E₂
E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶
E₁x = 10.95 × 10⁶ - 4.5 × 10⁶
E₁x = 6.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₁y = -E₁sinθ
E₁y = -12.24 × 10⁶ × sin(26.56)
E₁y = -12.24 × 10⁶ × 0.4471
E₁y = -5.5 × 10⁶ N/C.
Also, the magnitude of the electric field is given by:
Exy = √(E₁x² + E₁y²)
Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)
Exy = √4.225 × 10¹³ + 3.025 × 10¹³)
Exy = √(7.25 × 10¹³)
Exy = 8.5 × 10⁶ N/C.
Part B.We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:
E₂x = -E₂sinθ
E₂x = -4.5 × 10⁶ × sin(90)
E₂x = -4.5 × 10⁶ N/C.
Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:
E₂y = E₂cosθ
E₂y = -4.5 × 10⁶ × cos(90)
E₂y = 0 N/C.
How to calculate the net electric field at the observation location?The magnitude of the net electric field at B₃ is given by:
Eₙ = √(E₁x + E₂x)² + E₁y²)
Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)
Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)
Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)
Eₙ = √(3.65 × 10¹³)
Eₙ = 6.04 × 10⁶ N/C.
For the direction, we have:
Tanθ = x/y
Tanθ = 6.5/-5.5
Tanθ = -1.1818
θ = tan⁻¹(-1.1818)
θ = 49.76°.
Part C.The magnitude of the force on B₃ due to this net electric field is given by:
F = B₃ × Eₙ
F = 26 × 10⁻⁶ × 6.04 × 10⁶
F = 157.04 Newton.
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explain the use of low energy x-ray?
Answer:
The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.
*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
What is meant by significant figures
Answer:
The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.
Answer:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.
Explanation:
A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
a) Calculate its angular velocity in rad/s .
Express your answer using three significant figures.
b)What is the linear speed of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
c)What is the acceleration of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,
b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,
and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².
Calculation of angular velocity, linear speed & acceleration:
Provided that,
the diameter of the wheel = 0.35 m
So, the radius, r = 0.35/2 = 0.175 m
As 1 revolution = 2π rad
(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s
⇒ω = 272.13 rad/s
So, the angular velocity is 272.13 rad/s.
(b) The linear speed, v = r * ω
⇒v = 0.175 * 272.13
⇒v= 47.62 m/s
(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]
[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]
⇒[tex]a[/tex] = 12958.08 m/s²
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A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 revolutions before coming to rest.
Find the constant angular acceleration (in rad/s2) of the centrifuge.
______rad/s2
The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
Constant angular acceleration
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
ωf is the final angular velocity when the centrifuge stops = 0ωi is the initial angular velocity θ is angular displacementα is angular accelerationωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
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A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate of 0.21 m/s^2 and catches the ball after 3.8s. What was the dogs velocity when it caught the ball? *Assume here that running to the right is positive and running to the left is negative.
A- 7.96
B- 10.569
C- 6.782
D- 4.798
E- none
Answer:
D.-4.798m/s
Explanation:
Greetings !
Given values
[tex]u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec[/tex]
Solve for V of the given expression
Firstly, recall the velocity-time equation
[tex]v = u + at[/tex]
plug in known values to the equation
[tex]v = (4) + (0.21)(3.8)[/tex]
solve for final velocity
[tex]v = 4.792ms[/tex]
Hope it helps!
1. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain.
Answer:
(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?
No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum
(b)Is it possible for only one to be at rest after the collision?
Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.
why does a bus look red in the daylight?
Answer:
The red color in the sky at sunset (and sunrise) is due to an effect called Rayleigh scattering.
Explanation:
accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
formula for conductivity
Answer:
Thermal conductivity formula
k is the thermal conductivity (Wm -1 K -1)Q is the amount of heat transferred through the material (Js -1)A is the area of the body (m 2)ΔT is the temperature difference (K)Explanation:
In transverse waves, there is no _________ motion.
A. Perpendicular
B. Vertical
C. Horizontal
D. Up and down
In a transverse wave, the disturbance are carried forward and the particles are undergoing to and fro motion. Hence in transverse waves, there is no horizontal motion.
A wave can be described as a disturbance that travels through a medium from one location to the other location without transporting any matter.
There are two types of matter,
(i) Transverse wave
(ii) Longitudinal wave
In a transverse wave, the particles of the medium move in a direction perpendicular to the direction of propagation of the wave. That means, the particles undergo to and fro motion about their mean position and moves in the direction of the wave propagation.
Examples are:
Radio wavesLight wavesWater waves when a stone is dropped in the pondWhereas longitudinal waves undergo back and forth motion with respect to the mean position.
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I will run one mile in under 10 minutes 3 months from today.” is an example of a ___________ goal.
Long-term
Short-term
SMART
Both A and B
Answer:
long-term
Explanation:
usually a short term goal is able to be accomplished in a week or two. the question gives a 3 month time frame for the person to build up to the end goal.
The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = [tex]A_{0}[/tex][tex]e^{-btx}[/tex]cos(ωt) formula, where [tex]A_{0}[/tex] is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient.
(1) The period of the oscillator is 1 second.
(2) The damping coefficient is 0.93.
What is period of oscillation?
The period of oscillation is the time taken to make one complete cycle.
From the graph, the time taken to make one complete oscillation is 1 second.
Damping coefficientequation of the wave is given as;
y(t) = Ae^(-btx) cos(ωt)
at time, t = 0, y = 3.53.5 = Ae^(-0) cos(0)
3.5 = A x 1
A = 3.5 cm
at time, t = 1 cm, y = - 3cm-3 = 3.5e^(-bx) cos(ω)
-3/3.5 = e^(-bx) cos(ω)
-0.857 = e^(-bx) cos(ω)
-0.857 / cos(ω) = e^(-bx)
ln[-0.857 / cos(ω)] = -bx
ln[-0.857 / cos(ω)] / b = - x ---- (1)
at time, t = 2 cm, y = - 2cm-2 = 3.5e^(-2bx) cos(2ω)
-0.57 = e^(-2bx) cos(2ω)
ln[-0.57 / cos(2ω)] = -2bx
ln[-0.57 / cos(2ω)] /2b = - x ------(2)
solve (1) and (2)
ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b
-0.57 / cos(ω) = 2(-0.857 / cos(ω))
2(-0.857/cosω) = -0.57/cos2ω
-(2 x 0.857) / (-0.57) = cosω/cos 2ω
3 = cosω/cos 2ω
3(cos 2ω) = cosω
3(2cos²ω - 1) = cos ω
6cos²ω - 6 = cosω
6cos²ω - cosω - 6 = 0
let cosω = y
6y² - y - 6 = 0
solve the quadratic equation;
y = 1.1 or -0.92
cosω = -0.92
ω = arc cos(-0.92)
ω = 2.74 rad/s
From equation (1)
ln[-0.857 / cos(ω)] / x = -b ---- (1)
let x = 1
ln(-0.857/cos(2.74) = -b
-0.93 = -b
b = 0.93
Thus, the damping coefficient is 0.93.
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A light bulb in a camper’s flashlight is labeled 4.1 V, 0.40 A. Find the equivalent current if three of these light bulbs are connected in parallel to a standard C size 1.5 V battery.
Explanation:
first rule of parallel circuits : the voltage is the same in the whole circuit, no matter where we measure.
so, we have 1.5V at every bulb.
second rule of parallel circuits : the total current is the sum of the individual branch currents.
so, if the battery really allows it, then we have 3 times 0.4 A = 3×0.4 = 1.2 A as current in the circuit.
An object has a gravitational
potential energy of 750 Joules and is
sitting on a shelf 3.0 meters above
the ground. What is the object's
mass?
[?] kilograms
Answer:
potential energy = (mass × height × acceleration due to gravity.)
750 = mass × 3 × 10
hence, mass = 750/30 = 25 kg.
What is a small portable device that counts every step taken throughout the day?
A.stethoscope
B.pedometer
C.otoscope
D.thermometer
Answer: B
Explanation: i learned it last year
Pedometer is a small portable device that counts every step taken throughout the day. The correct option is (B).
A pedometer is a small portable device that counts the number of steps taken by an individual throughout the day. It is typically worn on the waist or carried in a pocket and uses a mechanism to detect body motion and count each step.
Pedometers are commonly used for tracking physical activity and encouraging individuals to achieve daily step goals as part of their fitness routines.
Therefore, Pedometer is a small portable device that counts every step taken throughout the day. Choose the option (B).
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The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres is 46622.72N
Calculation and Parameters( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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Have my equation written out but struggling to solve. Can someone help me solve!
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
How to find the missing current in a circuit
In this question we must make use of Kirchhoff's laws to find the values of the missing currents in the circuit presented in the picture. There are two rules according to Kirchhoff's laws:
The sum of currents found at nodes of circuits is always equal to zero.The net sum of voltages in a closed loop of a circuit is always equal to zero.Based on the information given by the picture, we have the following system of linear equations that describes the entire circuit:
i₃ = i₁ + i₂
- i₁ · R₁ + ε₁ - i₁ · r₁ - i₁ · R₅ + ε₂ - i₂ · (r₂ + R₂) = 0
ε₂ - i₂ · (r₂ + R₂) - i₃ · r₄ - ε₄ - i₃ · r₃ + ε₃ - i₃ · R₃ = 0
- i₁ - i₂ + i₃ = 0 (1)
(R₁ + r₁ + R₅) · i₁ + (r₂ + R₂) · i₂ = ε₁ + ε₂ (2)
(r₂ + R₂) · i₂ + (r₄ + r₃ + R₃) · i₃ = ε₂ + ε₃ - ε₄ (3)
If we know that R₁ = 5 Ω, r₁ = 0.10 Ω, R₅ = 20 Ω, r₂ = 0.50 Ω, R₂ = 40 Ω, r₄ = 0.20 Ω, r₃ = 0.05 Ω, R₃ = 78 Ω, ε₁ = 22 V, ε₂ = 49 V, ε₃ = 10.5 V and ε₄ = 33 V, then the currents flowing in the circuit are:
- i₁ - i₂ + i₃ = 0
25.1 · i₁ + 40.5 · i₂ = 71
25.1 · i₂ + 78.25 · i₃ = 26.5
The solution of this system is (i₁, i₂, i₃) = (- 10.852 A, 8.479 A, - 2.374 A). The negative signs indicate that real direction of the current is opposite than supposed.
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A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, y(x,t), is described by the following equation:
[tex]y(x,t)=0.0480[/tex]×[tex]sin(5.40x-128t)[/tex]
where x and y are in meters and the time is in seconds.
1. What is the wavelength of the wave?
2. What is the velocity of the wave? (Define positive velocity along the positive x-axis.)
3. What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)
The wavelength of the wave is 1.16m and the velocity is 23.64m/s.
To find the answer, we have to know more about the Transverse waves.
How to find different parameters of a wave?The displacement of the string as a function of position and time, y(x,t), when the wave traveling along a string lying along the x-axis is given as,[tex]y(x,t)=0.048sin(5.40x-128t)[/tex]
Comparing this with the general form of wave equation, we get,[tex]amplitude, a=0.048m\\wave vector, k=5.40\\angular frequency,w=128Hz[/tex]
We have to find the wavelength of the wave, for this, we have the expression as,[tex]k=\frac{2\pi }{wavelength} \\\\wavelength=\frac{2\pi }{k} =\frac{2*3.14}{5.40} \\\\wavelength=1.16m[/tex]
We have to find the velocity of the wave,[tex]v=frequency*wavelength\\2\pi f=w, thus,\\f=\frac{w}{2\pi } =\frac{128}{2*3.14}=20.38s^{-1}\\\\v=1.16*20.38=23.64m/s[/tex]
Thus, we can conclude that, the wavelength of the wave is 1.16m and the velocity is 23.64m/s
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a resistor wire of diameter 0.6cm has a resistivity of 1.0×00000010 ohms m. what length of the wire would be needed to make a 4 ohm resistor
The length of the wire is 1.12m
What is Resistance ?
Resistance is opposition to the current flowing in the circuit.
Given ,
D= 0.6m
R= 0.6/2m
R= 28Ω
rho = 1Ωm
R= rho ×l/A
where R is resistance , l is length ,A is area and rho is resistivity
By applying value and Solving we get,
l= 1.12m
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three objects are located in a coordinate system as shown below in figure a.find the centre of mass. how does the answer change if the object on the left is displaced upward by 1.00m and the object on the right is displaced downward by 0.500m (b) the object as point particles
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
What is the center of mass of a system of particles?A place at which the entire mass of the body or all the masses of a system of particles appears to be concentrated is known as the center of mass of a body or system of particles. According to physics, the center of mass is a location where the total of the weighted relative positions of the distributed mass's points in space equals zero.a)
The center of mass of a three-particle system is expressed as
[tex]r_{cm}=[/tex] [tex]$$r_{c m}=\frac{\sum_{i=1}^{3} m_{i} r_{i}}{\sum_{i=1}^{3} m_{i}} \Rightarrow \frac{m_{1} r_{1}+m_{2} r_{2}+m_{3} r_{3}}{m_{1}+m_{2}+m_{3}}$$[/tex]
When the system is only on [tex]$\mathrm{x}$[/tex]-axis (i.e. [tex]$\mathrm{y}=0$[/tex] )
[tex]$$\begin{aligned}x_{c m} &=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\x_{c m} &=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\x_{c m} &=0.5909 \mathrm{~m} \\\text { Therefore } r_{c m}=(0.5909 \mathrm{~m}, 0)\end{aligned}$$[/tex]
b)
When the two particles are shifted
[tex]\begin{aligned}&x_{c m}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}} \\&x_{c m}=\frac{(5.0 \times 0.5)+(2.0 \times 0)+(4.0 \times 1.0)}{5.0+2.0+4.0} \\&x_{c m}=0.5909 \mathrm{~m} \\&y_{c m}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}} \\&y_{c m}=\frac{(5.0 \times 1.0)+(2.0 \times 0)+(4.0 \times 0.5)}{5.0+2.0+4.0} \\&y_{c m}=0.6364 \mathrm{~m}\end{aligned}[/tex]
Therefore [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
The object as point particles is [tex]r_{cm} =0.5909 m, 0.6364 m.[/tex]
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PLS HELP 20 PIONTS!!
It is critical for scientists to be able to describe components of a system quantitatively. Explain why it is important to be able to describe a system quantitatively, using an example from your investigation about habitable worlds
A quantitative description of a system is important as it provides information about the carrying capacity of habitable worlds.
What is a quantitative description of the components of a system?Quantitative description refers to the description of a system which is focused on the numerical value of the properties or components of the system.
For example, a quantitative description of the components of a given habitat will be focused on the number of the individual species in the habitat. It will also be focused on the numerical value of the non-living components of the system and how such values affect the living components of the system numerically.
This information will then be used by scientists to see how modifications of the various quantitative components of the habitat will help to improve the chances of survival of species found in the habitat. This leads to such concepts as the carrying capacity of a habitat which is the maximum number of species that the available resources is a habitat can easily sustain.
In conclusion, a quantitative description is important to in investigations about the habitable world.
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A child in a boat throws a 5.90-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 25.0 kg and the mass of the boat is 38.4 kg. (Figure 1)
Calculate the velocity of the boat immediately after, assuming it was initially at rest.
Express your answer to three 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 box and negative value if the direction of the velocity is in the direction opposite to the velocity of the box.
-0.930 m/s is the velocity of the boat.
Given:Mass of child and boat , [tex]m_1[/tex] = (25.0 + 38.4 )kg
= 63.4 kg
Mass of the package, [tex]m_2[/tex] = 5.90 kg
Velocity of package thrown from boat , [tex]v_2[/tex] = 10.0m/s
[tex]v_1 =?[/tex]
Initial velocity v = 0 m/s
As the boat is at rest, [tex](m_1 + m_2) v=0[/tex]
According, to the law of conversation of momentum;
∴ Momentum before = Momentum after
[tex]( m_1 + m_2 ) v = m_1v_1 + m_2v_2\\0 = m_1v_1 + m_2v_2\\0 = 63.4 v_1 + 5.90*10.0\\63.4 v_1 = - 5.9\\v_1 = - 0.930 m/s[/tex]
Negative direction shows the velocity in the direction opposite to the motion of the package.
Therefore, -0.930 m/s is the velocity of the boat.
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*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude and direction of the force on a
−7.35 C charge at this location.
Magnitude _____N
Direction?
- north
-south
-east
-west
Answer:
Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)
Explanation:
If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].
The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:
[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].
An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:
[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves
The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.
Newton's Cannonball:Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.
Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.
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