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AysviL [449]
3 years ago
14

A wheel of mass 48 kg is lifted to a height of 0.8 m. How much gravitational potential energy is added to the wheel? Acceleratio

n Due to gravity is g = 9.8 m/s2.
A. 30.4 J
B. 3.1 J
C. 297.9 J
D. 11,321 J
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer: 376.32 J

Explanation: Gravitational potential energy is the energy possessed by a body by virtue of its position.

P.E=mgh

m= mass of the body = 48 kg

g= acceleration due to gravity= ]9.8m/s^2

h= height of body= 0.8 m

P.E=48kg\times 9.8m/s^2\times 0.8m=376.3J (1kgm^2s^{-2}=1 J)

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Answer:

Vector, in physics, a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude. Although a vector has magnitude and direction, it does not have position.

Explanation:

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3 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by p
nexus9112 [7]

Complete Question

An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?

A It should stay the same

B  It should be quadrupled.

C It should be quintupled

D It should be doubled.

E It should be tripled

Answer:

Option D is the correct option

Explanation:

Generally electric field is mathematically represented as

        E =  \frac{\sigma}{\epsilon_o}

Where \sigma is the charge per unit area (Charge density )

From the question we are told that \sigma is doubled hence the

     E =  \frac{2 \sigma }{\epsilon_o}    

Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density

8 0
3 years ago
Can anyone please tell me where these labels would go?
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6 0
3 years ago
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On a time line.
Sergeeva-Olga [200]

Answer:

D

Explanation:

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