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gregori [183]
4 years ago
5

An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the

arrow with a speed of 1.1m/s. The surface on which the target is sliding is friction-free. If the arrow embeds itself in the target, what will be the velocity of the target/arrow mass after the collision? Take the arrow's initial velocity direction as positive
Physics
1 answer:
Murrr4er [49]4 years ago
4 0

The velocity of the target and arrow after collision is 6.67m/s

<u>Explanation:</u>

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

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Friction between solids can be most accurately defined as the force that
ki77a [65]

Answer:

True

Explanation:

Friction between solids can be most accurately defined as the force that

opposes the sliding motion of two surfaces that are touching each other.

6 0
3 years ago
An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su
ra1l [238]

Answer:

Acceleration due to gravity will be g=17.3m/sec^2

Explanation:

We have given gauge pressure P = 3.8 atm = 3.8×101325 = 385035 Pa

Depth h = 24.3 m

Density \rho =1000kg/m^3

We have find the acceleration due to gravity at the surface of planet

We know that pressure is given by

P=\rho gh

So 385035=1000\times g\times 24.3

g=17.3m/sec^2

Acceleration due to gravity will be g=17.3m/sec^2

4 0
4 years ago
Select the correct answer.
BaLLatris [955]

It is due to the excess stress.

6 0
3 years ago
What should scientists do after completing a scientific investigation?
ivanzaharov [21]
B sounds most logical
7 0
3 years ago
A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upwa
Ksju [112]

The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.

The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2

=> v^2 = 11556/5

=> v = 48.07

<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>

3 0
3 years ago
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