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Vanyuwa [196]
3 years ago
15

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m

oving in the same direction at a speed v2.
(a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2? vf =

(b) How much kinetic energy is lost in the collision? Answer in terms of M, v1, and v2. ?KE =
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

Explanation:

Given

mass of each railroad is M

first railroad is moving with velocity v_1 and other two's with v_2

conserving Momentum

Mv_1+2Mv_2=3M v_f

where v_f is the final velocity of combined system

v_f=\frac{v_1+2v_2}{3}

(b)Initial Kinetic Energy

=\frac{1}{2}Mv_1^2+\frac{1}{2}2Mv_2^2

=\frac{1}{2}Mv_1^2+Mv_2^2

Final Kinetic Energy

=\frac{1}{2}3Mv_f^2=\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

Loss in Kinetic Energy

=\frac{1}{2}Mv_1^2+Mv_2^2-\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

=M(v_1-v_2^2)^2

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3 years ago
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A parking lot is going to be 60 m wide and 240 m long. what dimensions could be used for a scale model of the lot?​
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Answer:

it is 20cm x 80cm

Explanation:

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A fairground ride consists of a large vertical drum that spins so
expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}

<u>v = 3.84 m/s</u>

6 0
3 years ago
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o
masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, F_{f2} = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = F_{f2} × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27

Frictional \ force, F_{f2} = \left (981 -  \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8}  \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times  0.25 \right) \approx 219.92

The frictional force on the block W₂, F_{f2} ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ F_{w2} = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + \mathbf{F_{w2}}

The frictional force from the ground, \mathbf{F_{f1}} = N×μ + \mathbf{F_{f2}} = P

Where;

P = The horizontal force applied to the block

P = (W₁ + \mathbf{F_{w2}}) × μ + \mathbf{F_{f2}}

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0
3 years ago
In this first example of constant accelerated motion, we will simply consider a car that is initially traveling along a straight
algol13

Answer:

v_{f} =25m/s

Explanation:

Kinematics equation for constant acceleration:

v_{f}  =v_{o} + at=15+2*5=25m/s

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