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Alexxx [7]
3 years ago
7

In an inelastic collision involving an isolated system, the final total momentum is

Physics
2 answers:
Aleksandr-060686 [28]3 years ago
8 0

Answer: B) exactly the same as the initial momentum.

Explanation:

An inelastic collision occurs when the elements that collide remain together after the collision, and althogh the kinetic energy is not conserved because is transformed into other kinds of energy (thermal energy, for example), the linear momentum does.  

This means the initial momentum before the collision will be equal to the final momentum after the collision:

p_{o}=p_{f}

Sindrei [870]3 years ago
6 0

Answer:

exactly the same as the initial momentum.

Explanation:

There are two types of collisions :

1. Elastic collision

2. Inelastic collision

In an inelastic collision, the linear momentum before and after the collision remains conserved while the kinetic energy is not conserved. Some of the energy gets lost in the form of heat, light, sound etc.

So, In an inelastic collision involving an isolated system, the final total momentum is exactly the same as the initial momentum.

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Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

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An airplane is moving at 350 km/hr. If a bomb is
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a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

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y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

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Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

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