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3 years ago

When a person steps forward out of a small boat onto a dock, the boat recoils backward in the water. why does this occur??

Physics

2 answers:

Serjik [45]3 years ago

6 0

Answer:

Due to conservation of momentum

Explanation:

We know that if there is no any force acting on the system then the total linear momentum of system will remain conserve.We know that linear momentum given as

P = m x v

When person steps forward then velocity of person changes it means that the linear momentum also changes to balance this change in linear momentum the boat will move away from the bank. Because the linear momentum should be conserve.

bixtya [17]3 years ago

4 0

Here is the correct answer of the given question above. When a person steps forward out of a small boat onto a dock, the boat recoils backward in the water and this occurs because the total momentum of the system is conserved. Hope this helps.

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10 points

grandymaker [24]

The answer is B tell me if I am wrong.

6 0

2 years ago

The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />

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3 years ago

One of the world's largest Ferris wheels, the Cosmo Clock 21 with a radius of 50.0 m is located in Yokohama City, Japan. Each of

STatiana [176]

Answer:

a = 0.55 m / s²

Explanation:

The centripetal acceleration is given by the relation

a = v² / r

angular and linear velocities are related

v = w r

we substitute

a = w² r

In the exercise they indicate the angular velocity w = 1 rev/min, let's reduce to the SI system

w = 1 rev / min (2pi rad / 1rev) (1min / 60s) = 0.105 rad/ s

let's calculate

a = 0.105² 50.0

a = 0.55 m / s²

4 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$