Supposing velocity is speed and direction, and momentum is mass*velocity, if the velocity increases by a factor of 5, then so should the momentum regardless of the whole space thing.
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Answer:

Explanation:
We can use the equation for the speed

where x is the distance and t the time. In this case we know that the time spent was 2 hours and the distance was 150km. By replacing we have

I hope this useful for you
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Answer:
The acceleration of Abbie is half of the Zak's.
Explanation:
The centripetal acceleration of an object on a circular path is given by :

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let
is distance of Abbie from the merry-go-round and
is distance of Zak's from the merry-go-round. Acceleration of Abbie is :
...... (1)

Acceleration of Zak's is :
.......(2)

Dividing equation (1) and (2) we get :

So, the acceleration of Abbie is half of the Zak's.
The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
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