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kondor19780726 [428]
3 years ago
14

You are walking on a moving walkway in the airport. The length of the walkway is 59.1 m. If your velocity relative to the walkwa

y is 2.35 m/s, and the walkway moves with a velocity of 1.85 m/s, how long will it take you to reach the other end of the walkway
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

14.1seconds approx

Explanation:

Given data

Distance= 59.1m

Your velocity= 2.35m/s

Walkway velocity= 1.85m/s

Total velocity= 2.35+1.85= 4.2m/s

We know that

Velocity= distance/time

time= distance/velocity

substitute

time= 59.1/4.2

time= 14.07

time=14.1seconds approx

Hence the time is 14.1seconds approx

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3 years ago
A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg l
melisa1 [442]

Answer:

v_f=-0.29\frac{m}{s}

Explanation:

The principle of conservation of momentum, states that if the sum of the forces acting on a system is null, the initial total momentum of the system before a collision equals the final total momentum of the system after the collision. The collision is completely inelastic, which means that the players remain stick to each other after the collision:

p_i=p_f\\m_1v_1+m_2v_2=(m_1+m_2)v_f\\v_f=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\v_f=\frac{(110kg)4.5\frac{m}{s}+150kg(-3.8\frac{m}{s})}{(110kg+150kg)}\\v_f=-0.29\frac{m}{s}

5 0
3 years ago
As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65
kaheart [24]

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

\theta = Angle at which the force is being applied = 65^{\circ}

Horizontal component of force is given by

F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}

The horizontal component of the force acting on the crate is 19.01 N.

4 0
2 years ago
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

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3 years ago
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The animal that is hunted and consumed is considered the prey

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3 years ago
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