Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
The longer you spend reading and thinking about this question,
the more defective it appears.
-- In each case, the amount of work done is determined by the strength
of
the force AND by the distance the skateboard rolls <em><u>while you're still
</u></em>
<em><u>applying the force</u>. </em>Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.<em>
</em>
-- We know that the forces are equal, but we don't know anything about
how far each one rolled <em>while the force continued</em>. All we know is that
one force must have been removed.
-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.
-- How far did that one roll while you were still pushing it ?
-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?
-- Did each skateboard both roll the same distance while you continued pushing it ?
I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
Answer:
The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
<span>A baseball speeds up as it falls through the air.
Yes. Forces on the balloon are unbalanced.
The balloon is speeding up, so we know that the downward force
of gravity is stronger than the upward force of air resistance.
A soccer ball is at rest on the ground.
No. The ball is not accelerating, so we know that the forces on it
are balanced.
The downward force of gravity on the ball and the upward force
of the ground are equal.
An ice skater glides in a straight line at a constant speed.
No. The skater's speed and direction are not changing, so he is not
accelerating. That tells us that the forces on him are balanced.
A bumper car hit by another car moves off at an angle.
Yes. The direction in which the car was moving changed.
That's acceleration, so we know that the forces on it are unbalanced,
at least at the moment of impact.
A balloon flies across the room when the air is released.
Yes. The balloon was not moving. But when the little nozzle was
opened, it started to zip around the room. So its speed changed.
And, as it goes bloozing around the room, its direction keeps changing too.
There's a whole lot of acceleration going on, so we know the forces on it
are unbalanced.</span>
