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zheka24 [161]
3 years ago
10

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

nument. A nickel with the mass of 0.005 kg is in her shirt pocket. What is the kinetic energy (KE) of the nickel in her shirt pocket at the top of the monument?
Physics
1 answer:
Mnenie [13.5K]3 years ago
7 0

Answer:

Your answer is: K.E = 8.3 J

Explanation:

If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.

K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula  ↓

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

So your answer is 8.2908 but if you round it is K.E = 8.3

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Based on this cell from the periodic table, which statement is true for the
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Answer: B

Explanation:

The top number is its atomic number, also known as the number of protons in the atom.

The bottom number is the atomic mass, which is the sum of the number of protons and neutrons.

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Does a volcano contain a medium? Does the energy travel through anything- the transverse waves?
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3 years ago
4. Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 7.3 times tha
HACTEHA [7]

Answer:

Explanation:

Acceleration of particle A is 7.3 times the acceleration of particle B.

Let the acceleration of particle B is a, then the acceleration of particle A is

7.3 a.

Let the period of particle A is T and the period of particle B is 2.5 T.

Let the radius of particle A is RA and the radius of particle B is RB.

Use the formula for the centripetal force

a=r\omega ^{2}=r\times \frac{4\pi^{2}}{T^{2}}

So, r = a\frac{T^{2}}{4\pi^{2}}

The ratio of radius of A to the radius of B is given by

\frac{R_{A}}{R_{B}}=\frac{a_{A}\times T_{A}^{2}}{a_{B}\times T_{B}^{2}}

\frac{R_{A}}{R_{B}}=\frac{7.3 a\times T^{2}}{a\times 6.25T^{2}}

RA : RB = 1.17

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3 years ago
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A 900 kg vehicle moves around a curve with an incline of 20\circ∘ at a speed of 12.5 m/s. If the curve has a radius of 50 meters
valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

The incline of the curve, θ = 20°

The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}

Where;

θ = The angle of inclination of the road = 20°

N = The normal force experienced by the car

m = The mass of the car = 900 kg

v = The velocity with which the car is moving = 12.5 m/s

R = The radius of the curve around which the vehicle moves = 50 m

\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})}  = 8223.1998754586828969046217875927

The normal force experienced by the car = N ≈ 8223.2 N.

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2 years ago
Jessica pulls on a washing machine with 200 N of force. Mark helps out and pulls on the washing machine in the same direction wi
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Answer:

600N

Explanation:

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