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Fynjy0 [20]
3 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, aft

er it has fallen 22 m:
How much time has passed (in s)?
what is the speed at this distance (in m/s)?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Since the object is dropped from some height so its initial speed must be zero

acceleration of the object is due to gravity

so we can use kinematics to find the time it will take to drop by x = 22 m

\delta x = v_i * t + \frac{1}{2}at^2

22 = 0 + \frac{1}{2}*9.8*t^2

t = 2.12 s

Now the speed after 2.12 s will be given as

v_f = v_i + at

v_f = 0 + 9.8 * 2.12

v_f = 20.8 m/s

so above is the speed and time

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Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/
madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

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8 0
3 years ago
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Rama09 [41]

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5 0
3 years ago
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I am Lyosha [343]
I think Option B, i’m sorry if that’s wrong.
7 0
3 years ago
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Andreyy89

Answer:

Plan B.

Because flexibility is best improved by stretching.

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Which of the following would have an effect on an object's mass?
ICE Princess25 [194]
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8 0
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