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Fittoniya [83]
3 years ago
8

A cart of mass m moving right at speed v with respect to the track collides with a cart of mass 0.7m moving left.

Physics
2 answers:
bagirrra123 [75]3 years ago
6 0

Answer:

10/7 * V.

Explanation:

Momentum = mass * velocity

Given:

Mass of cart 1 = M1

Mass of cart 2 = M2

= 0.7M1

Initial velocity of cart 1 = Vi1

= V

Initial velocity of cart 2 = Vi2

Final velocity = V

= 0

Initial momentum of cart 1 + initial momentum of cart 2 = total mass of both cart * velocity

M1 * Vi1 + M2 * Vi2 = (M1 + M2) * V

= M * V - (0.7M * Vi2) = (M + 0.7M) * 0

= MV - 0.7M * Vi2 = 0

Vi2 = 10/7 * V.

Zolol [24]3 years ago
5 0

Answer:

10v / 7

Explanation:

Using the conservation law of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m v - 0.7 m v₁ =  ( 0.7 m + m) 0 m/s since the cart stuck together after collision. taken right to be positive and left to be negative

m v - 0.7 m v₁ = 0

- 0.7 m v₁ = -m v

v₁ = -m v / - 0.7 m = 10v / 7

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
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Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

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P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

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3 years ago
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Answer:

8 V

Explanation:

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3 years ago
Read 2 more answers
A 1500 kg truck traveling at 80 km/h collides with another car of mass 1000 kg traveling at 30 km/h in the same direction. The t
Likurg_2 [28]

Answer:

Their speed immediately after the collision is 60 km/h

Explanation:

The magnitude that characterizes the state of motion of a body is called the quantity of motion. The momentum of a particle of mass m moving with velocity v is defined as the product of mass and velocity:

p=m*v

Newton's second law states that to accelerate an object you must apply a force to it. In other words, to change the moment of an object, an impulse must be applied to it, an impulse that is produced by a force. In both cases there is an external agent that exerts the force or the impulse, the internal forces are not considered. When the force  net is zero, so the net momentum is zero, and therefore there is no change in total momentum. Then it can be stated that if a net force is not exerted on a system, the total moment of the system cannot change.

In summary, the principle of conservation of linear momentum, also known as the principle of conservation of momentum, states that if the resultant of the forces acting on a body or system is zero, its momentum remains constant in time.

Then before collision, the momentum of truck is  1,500 kg*80 km/h = 120,000 kg.km/h  and the momentum of car is  1,000 kg* 30 km/h = 30,000 kg.km/h

So, the total momentum before collision is  120,000 kg*km/h+ 30,000 kg*km/h = 150,000 kg.km/h

The two cars stick together after the collision. This means that the speed V of both is the same. Therefore, the momentum after collision is  

V*(1,500 kg + 1,000 kg)  = V*2,500 kg

Applying the law of conservation of momentum, that it states that the total momentum of two objects before collision is equal to the total momentum of the two objects after collision., you get:

150,000 kg.km/h= V*2,500 kg

Solving:

V=\frac{ 150,000 kg.km/h}{2,500 kg}

V= 60 km/h

<u><em>Their speed immediately after the collision is 60 km/h</em></u>

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