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nydimaria [60]
3 years ago
8

A medicine ball is dropped from a height of 1 m. It strikes the ground with an energy of 50 J. What is the mass of the medicine

ball?
Physics
1 answer:
Otrada [13]3 years ago
8 0

Answer:

5 kg

Explanation:

PE = KE

mgh = KE

m (9.8 m/s²) (1 m) = 50 J

m = 5.1 kg

Rounded to one significant figure, the mass is 5 kg.

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A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of
kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0
3 years ago
In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

\omega =2.55 rad/sec

8 0
3 years ago
Eric has a mass of 19.0 kg on the earth. What is Eric's weight on earth? What is Eric's weight on Mars? where the acceleration o
Tom [10]

Answer:

Weight on Earth = We = 186.2 N

Weight on Mars = Wm = 70.94 N

Explanation:

The weight of an object is defined as the force applied on the object by the gravitational field. The magnitude of weight is given by the following formula:

W = mg

were,

W= Weight of Eric

m = mass of Eric

g = acceleration due to gravity

ON EARTH:

W = We = Eric's Weight on Earth = ?

m = Eric's Mass on Earth = 19 kg

ge = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

We = (19 kg)(9.8 m/s²)

<u>We = 186.2 N</u>

<u></u>

ON MARS:

W = Wm = Eric's Weight on Mars = ?

m = Eric's Mass on Mars = 19 kg

gm = acceleration due to gravity on Mars = 0.381(ge) = (0.381)9.8 m/s² = 3.733 m/s²

Therefore,

Wm = (19 kg)(3.733 m/s²)

<u>Wm = 70.94 N</u>

3 0
3 years ago
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