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
It’s 4 because a coiled springs is closely spaced then widen
Answer:
Bob's angular speed is the same as that of lily
Explanation:
Because for a carousel the angular speed remains the same since velocity at center and edge are the same
Answer:
The answer is D
Explanation:
I'm too lz to explain everything.
sorry.
1) 29.8 C
At the beginning, the metal is at higher temperature (70.4 C) while the water is at lower temperature (23.6 C). When they are put in contact, the metal transfers heat to the water, until they reach thermal equilibrium: at thermal equilibrium the two objects (the metal and the water have same temperature). Therefore, since the temperature of the water at thermal equilibrium is 29.8 C, the final temperature of the metal must be the same (29.8 C).
2) 6.2 C
The temperature change of the water is given by the difference between its final temperature and its initial temperature:

where

Substituting into the formula,

And the positive sign means that the temperature of the water has increased.
3) -40.6 C
The temperature change of the metal is given by the difference between its final temperature and its initial temperature:

where

Substituting into the formula,

And the negative sign means the temperature of the metal has decreased.