The final velocity is 2.7 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.
Therefore we can write:
where:
is the mass of the putty
is the initial velocity of the putty (we take its direction as positive direction)
is the mass of the ball
is the initial velocity of the ball (at rest)
is the final combined velocity of the two putty+ball
Re-arranging the equation and substituting the values, we find the final combined velocity:
And the positive sign indicates their final direction is the same as the initial direction of the putty.
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Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
The force is the same
Explanation:
The force per meter exerted between two wires carrying a current is given by the formula

where
is the vacuum permeability
is the current in the 1st wire
is the current in the 2nd wire
r is the separation between the wires
In this problem

Substituting, we find the force per unit length on the two wires:

However, the formula is the same for the two wires: this means that the force per meter exerted on the two wires is the same.
The same conclusion comes out from Newton's third law of motion, which states that when an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A (action-reaction). If we apply the law to this situation, we see that the force exerted by wire 1 on wire 2 is the same as the force exerted by wire 2 on wire 1 (however the direction is opposite).
Answer:
sure, do you have more detailed of the assignments...?
Explanation:
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