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Evgesh-ka [11]
3 years ago
9

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car

initially has a total mass (car and contents) of 150 kg and is traveling east with a velocity of magnitude 4.60 m/s. Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
Part A : An object with a mass of 21.0 kg is thrown sideways out of the car with a speed of 2.40 m/s relative to the car's initial velocity.
Part B : An object with a mass of 21.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Part C : An object with a mass of 21.0 kg is thrown into the car with a velocity of 6.00 m/s relative to the ground and opposite in direction to the initial velocity of the car.
Physics
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

Part a)

v = 4.60 m/s

Part b)

v_{1f} = 5.24 m/s

Part c)

v = 3.3 m/s

Explanation:

Part a)

Object is thrown sideways

since the railroad car is moving on the track so there is no change in the momentum in the direction of motion of the car

so final speed of the car will not change

v = 4.60 m/s

Part b)

By momentum conservation we can write

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

150(4.60) = 150 v_{1f} - 21(4.60)

v_{1f} = 5.24 m/s

Part c)

Again by momentum conservation we have

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f

150(4.60) - 21(6.00) = (150 + 21) v

v = 3.3 m/s

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Surgical needles are classified according to their shape; a size of circle shaft within an x,y axis; their point; and their eye.
gogolik [260]

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3 0
3 years ago
A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

8 0
3 years ago
Agent Bond is standing on a bridge, 17 m above the road below, and his pursuers are getting too close for comfort. He spots a fl
faust18 [17]

Answer:

2 poles

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Now, the bed of the truck is 1.5 m above the road.

It means the distance hewill need to fall = 17 - 1.5 = 15.5 m

acceleration due to gravity = 9.8 m/s²

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Thus, to find the time, we will use the equation;

s = ut + ½at²

Plugging in the relevant values gives;

15.5 = 0 + ½9.8t²

Multiply through by 2 to give;

15.5 × 2 = 9.8t²

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t² = 31/9.8

t = √(31/9.8)

t = 1.78 sec

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Thus; The truck will pass the poles at a rate of; Velocity/distance between poles = 28/22 = 1.273 poles per second

Since, we got the time to be 1.78 seconds, then we can find the number of poles by multiplying it with the rate of poles per seconds.

Thus;

Number of poles = 1.78 second × 1.273 poles/ seconds = 2.27 poles

This is approximately 2 poles

6 0
3 years ago
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