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Fudgin [204]
3 years ago
13

A child is perched perilously above a 150m ravine. The child has a mass of 45kg. How much

Physics
1 answer:
Finger [1]3 years ago
6 0

Answer:

<em>U = 66,150 J</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy is the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8\ m/s^2.

The child of mass m=45 Kg is perched above a h=150 m ravine. His gravitational potential energy is:

U=45~Kg\cdot 9.8\ m/s^2\cdot 150\ m

U = 66,150 J

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A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac
7nadin3 [17]

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

8 0
2 years ago
A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of
SCORPION-xisa [38]

Answer:

a).β=0.53x10^{-3} T

a).β=0.40 x10^{-4} T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

\beta =\frac{u_{o}*I*N}{2\pi*r}

a).

N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A}  \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T

b).

The distance is the radius add the cross section so:

r_{1}=15cm+5cm\\r_{1}=20cm

r_{1} =20cm*\frac{1m}{100cm}=0.20m

\beta =\frac{u_{o}*I*N}{2\pi*r1}

\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T

3 0
3 years ago
ASAP!! please What is the answer? MCQ. Thank you
sdas [7]

Answer:

Id say C

Explanation:

The hypotenuse for c is up and to the right so the unit vectors show that

8 0
3 years ago
Read 2 more answers
13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.
julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

7 0
3 years ago
Calculate the energy gained by an ice block in the following experiment.
MAXImum [283]

The heat is exchanged when two different temperature objects come in contact. The energy gained by an ice block is 2.3 Joules.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the object.

A 7g block of ice was added to a coffee cup full of 103.4 grams of water. The water had an initial temperature T₁ = 24.5 C and a final temperature T₂ = 19.2 C after all the ice had melted.

Heat lost by water = Heat gained by ice

Qgain = ms(T₂ -T₁ )

Substituting the value for mass of water m =103.4 g= 0.1034 kg , specific heat of water s = 4.18 kJ/kg and temperature values, we get

Qgain = 0.1034 x 4.18 x (24.5 - 19.2)

Qgain = 2.3 Joules

Thus, the energy gained by an ice block is 2.3 Joules.

Learn more about temperature.

brainly.com/question/20459283

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7 0
2 years ago
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