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Aleks04 [339]
3 years ago
15

6) A ball free falls from the top of the roof for 5 seconds. How far did it fall? What is its final

Physics
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

distance fallen = 122.5 m

speed = 49 m/s

Explanation:

Δd = vi*t - 0.5at^2

Δd = 0 - 0.5(9.8)(5^2)

Δd = -122.5 (i.e. 122.5m down)

a = (vf - vi) / t

9.8 = (vf - 0) / 5

vf = 49

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En un barrio un estudiante observa un roedor en medio de dos postes, los cuales están comunicados por un solo cable de fibra ópt
castortr0y [4]

Answer:

7 Newton

Explanation:

Dado

Longitud de la cuerda = 50 m

El cable se dobla en 0,058 m.

Masa de roedor = 350 gramos = 0,35 kg

T = m * a + m × v2 / r

Sustituyendo los valores dados obtenemos

T = 0,35 (10 + 10)

T = 0,35 * 20

T = 7 Newton

4 0
3 years ago
Which type of experiment involves changing only one variable at a time
BlackZzzverrR [31]
Controlled Experiment 

6 0
3 years ago
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A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h
Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = \frac{dr}{dt} X 12 = 4 \frac{ft}{sec}  X 12 sec = 48 ft

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

\frac{dA}{dt} = 2\pi (48)(4)

    dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

7 0
3 years ago
Suppose you took a trip to the moon. Write a paragraph describing how and why your weight would change. Would your mass change t
noname [10]
Your weight would change but not your mass, the moon has less gravity so therefore you are going to be lighter :-)
5 0
3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

Which means KE = m·g·h

We can solve for h:

h = KE / m·g
   = 20 / (0.15 · 9.8) 
   = 13.6m

The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
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