For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
Answer:48 V
Explanation:
Given
Three charged particle with charge
Electric Potential is given by
Distance of 
from 
similarly 
Potential at is
![V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]](https://tex.z-dn.net/?f=V_%7Bnet%7D%3Dk%5B%5Cfrac%7Bq_1%7D%7Bd_1%7D%2B%5Cfrac%7Bq_2%7D%7Bd_2%7D%2B%5Cfrac%7Bq_3%7D%7Bd_3%7D%5D)
![V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}](https://tex.z-dn.net/?f=V_%7Bnet%7D%3D9%5Ctimes%2010%5E9%5B%5Cfrac%7B50%7D%7B10%7D-%5Cfrac%7B80%7D%7B12%7D%2B%5Cfrac%7B70%7D%7B10%7D%5D%5Ctimes%2010%5E%7B-9%7D)
The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:
Vf = Vi + at
where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
Substituting the values gives:
30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
Answer:
the motion of the coin taping the balloon is the balloon squshing down
