Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Explanation:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Answer:
5,000J
Explanation:
Work = Force x Distance
Distance back and forth is canceled out, so either the answer is + or -
5.0m + 5.0m = 10.0m
500N x 10.0m = 5,000J
Answer:
Coefficient of friction = 0.5
Explanation:
Given:
Mass of box = 5 kg
Force applied = 20 N
Acceleration = 2 m/s²
Find:
Coefficient of friction
Computation:
Friction force = Mass x Acceleration.
Friction force = 5 x 2
Friction force = 10 N
Coefficient of friction = Friction force / Force applied
Coefficient of friction = 10 / 20
Coefficient of friction = 0.5