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guajiro [1.7K]
3 years ago
13

magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t

o be at rest after the collision, how fast must the less massive cart move compared to the more massive cart
Physics
1 answer:
scoray [572]3 years ago
8 0

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

                             P_i = P_f

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

                             P_i = P_f = 0

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

                             m*u - 2*m*v = 0

Where,

                 ( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

                          m*u = 2*m*v

                              u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

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Answer:

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Explanation:

Given that,

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We need to calculate the greatest friction force

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Put the value into the formula

F = 0.77\times85\times9.8

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Using formula of newton's second law

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{ 641.41}{109000}

a=0.0059\ m/s^2

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0
3 years ago
Find τf, the torque about point p due to the force applied by the achilles' tendon.
Luba_88 [7]
The formula for the torque is
<span>τf = p F
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3 0
3 years ago
A 125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
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Answer:

201.6 N

Explanation:

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r = radius of the disk = 1.50 m

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t = time interval = 2 s

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Using the equation

w = w₀ + α t

4.296 = 0 + 2α

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Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

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Torque equation for the merry-go-round is given as

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4 0
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The bar graph shows energy data taken from a roller coaster at a theme park. analyze the data and assess its validity. 3-5 sente
hichkok12 [17]

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The data shown by the bar graph is "valid".

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From the data given in the bar graph:

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From the data given in the bar graph:

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Hence,

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Learn more about the law of conservation of energy here:

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The attached picture explains the law of conservation of energy.

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lord [1]

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Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

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