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ahrayia [7]
4 years ago
14

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. Wha

t is its vertical displacement (in m) after 6 s?
Physics
1 answer:
Snowcat [4.5K]4 years ago
3 0

Answer:

176.4 m

Explanation:

U = 0, t = 6s, g = 9.8 m/s^2

Use second equation of motion

H = ut + 1/2 gt^2

H = 0 + 0.5 × 9.8 × 6 × 6

H = 176.4 m

It is the displacement from the point of dropping of object.

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Id say the rubber band
3 0
3 years ago
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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
An experiment yielded the above temperature and time information. what is the freezing point of the material in this experiment
Dmitry [639]
The answer is 0 degrees Celsius (0°C). It will be where the line flat lines the first time.  The second time would be the boiling point.  An experiment yielded the above temperature and time information. The freezing point of the material in this experiment if the material is a solid at time zero is 0 degrees Celsius (0°C) .
7 0
3 years ago
abody starts from rest and accelerate uniformly at 8m/s/s.calculate the distance travelled by the body in 10s? ​
vagabundo [1.1K]

Explanation:

using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.

S=ut+½gt² ("ut" term will cancel because u=0).

=> S= ½gt²

=>S = ½×8×10²

=>S = 4×100

=>S = 400m .

Therefore, the distance traveled by the body in 10s is 400m.

hope this helps you.

6 0
2 years ago
URGENT
Advocard [28]

Answer:

give \\ mass(m) = 1350kg  \\ acceleration(a) = 1ms {}^{ - 2}  \ \\ sln\ \\  from \: our \: formula \\  \: f = ma \\1350kg \times 1ms {}^{ - 2}  \\ f = 1350newton

the force you applied to your car =1350N

5 0
3 years ago
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