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photoshop1234 [79]
3 years ago
8

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

from one end of the car to the other in the direction of motion, with speed 2.0 m/s with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?
Physics
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

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Hello! :)

The focal length of the lens tells you how far away from the lens a focused image is created, if light rays approaching the lens are parallel. A lens with more “bending power” has a shorter focal length, because it alters the path of the light rays more effectively than a weaker lens. Most of the time, you can treat a lens as being thin and ignore any effects from the thickness, because the thickness of the lens is much less than the focal length. But for thicker lenses, how thick they are does make a difference, and in general, results in a shorter focal length.

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5 0
3 years ago
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro
ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

T=2\pi\sqrt{\frac{l}{g} }

angular frequency ω = 2π / T

= \omega=\sqrt{\frac{g}{l} }

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 (  ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ  = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0
3 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
According to the video, what do Electrical Engineers usually study in college? Check all that apply.
Dmitry [639]

Answer:

Math & Science

Explanation:

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6 0
3 years ago
Read 2 more answers
A student wants to approximate the amount of work that the force due to gravity does on the student as the student walks up a se
Inessa [10]

Answer:

Options A and D

Explanation:

In this question the student needs to collect these measurements in order to approximate the work done

A. The mass of the student

D. The final vertical height above the initial vertical position.

Workdone = mgh

m = mass

g = gravity = 9.8m/s²

h = vertical height between the initial and the final positions.

The vertical height has to be known as gravity only acts straight down.

3 0
3 years ago
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