A ball bounces on the ground. How do the ball and the ground act on each other?(1 point)
2 answers:
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a) 2.64 s
We can solve this part of the problem by using the following SUVAT equation:
where
s is the displacement of the stone
u is the initial velocity
t is the time
a is the acceleration
We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:
- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.
b) 10.4 m/s
The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:
where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
Substituting t = 2.64 s, we find the final velocity of the stone:
where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.
c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
We have:
s = -105 m (vertical displacement of the package, downward so negative)
u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)
a = g = -9.8 m/s^2
Substituting into the equation,
Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after
t = 4.11 seconds.
Answer:
t = 23.9nS
Explanation:
given :
Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m
distance d= 1mm=> 0.001
resistor R= 975 ohm
Capacitance can be calculated through the following formula,
C = (ε0 x A )/d
C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001
C = 17.7 x 10^-12 (pico 'p' = 10^-12)
C = 17.7pF
the voltage between two plates is related to time, There we use the following formula of the final voltage
Vc = Vx (1-e^-(t/CR))
75 = 100 x (1-e^-(t/CR))
75/100 = (1-e^-(t/CR))
.75 = (1-e^-(t/CR))
.75 -1 = -e^-(t/CR)
-0.25 = -e^-(t/CR) --->(cancelling out the negative sign)
e^-(t/CR) = 0.25
in order to remove the exponent, take logs on both sides
-t/CR = ln (0.25)
t/CR = -ln(0.25)
t = -CR x ln (0.25)
t = -(17.7 x 10^-12 x 975) x (-1.38629)
t = 23.9 x
t = 23.9ns
Thus, it took 23.9ns for the potential difference between the deflection plates to reach 75 volts