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Darya [45]
3 years ago
11

Terry dropped a pebble from the top of his apartment building. The pebble impacted the ground after falling for 2.2 seconds. Cal

culate the final velocity of the pebble at impact with the street below. -4.45 m/s -21.5 m/s -23.7 m/s -9.8 m/s​
Physics
1 answer:
Contact [7]3 years ago
6 0

The final velocity of the pebble as it hits the ground is 21.5 m/s.

We have to use the equation;

v = u + gt

Where;

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Since the pebble was dropped from a height, u =0 m/s, hence,

v = gt

v = 9.8 ms-2 × 2.2 s

v = 21.5 m/s

Learn more about velocity: brainly.com/question/8898885

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A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
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Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
Identify the frequency used by these broadcast media.<br><br> FM<br> AM<br> Television
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Answer:

AM radio = 540 - 1650 KHz

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Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

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1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

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Answer:

3 m/s

Explanation:

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