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notka56 [123]
3 years ago
10

An object of mass 5 m is lifted to a height of 5 h above the floor. What is the potential energy the object gained in this proce

ss (in the unit of mgh)? The potential change is
Physics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Potential energy, E = 25 mgh

Explanation:

Given that,

Mass of the object, m = 5 m

It is lifted to a height of, h = 5 h

It is required to find the potential energy gained by the object. It is given by the product of object's mass, acceleration due to gravity and heigh above ground. It is given by :

E=m\times g\times h

E=5\ m\times g\times 5\ h

E = 25 mgh

So, the potential energy gained by the object is (25 mgh). Hence, this is the required solution.

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A soccer ball was kicked off the roof of a building going
WINSTONCH [101]

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

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3 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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