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fiasKO [112]
4 years ago
5

Would thermal energy be greater at 0°C or 48°F?

Physics
1 answer:
Alik [6]4 years ago
5 0
If you know 0 C= 32 F, it is greater at 48 F
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hen a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant
Sladkaya [172]

Answer:

Incomplete question

Complete question:

a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with mass 44kg , the acceleration of the object is 4m/s². If the same force acts upon another object whose mass is 11kg what is this object's acceleration?

Answer: 8m/s²

Explanation:

From the statement we deduced that acceleration varies inversely with mass where force was kept constant.

Therefore,

F/m = a or F = ma

For the first statement, substituting the mass and acceleration gives:

F = 44 x 4 = 88N

Applying the force above to the second mass gives us:

a = 88/11 = 8m/s²

3 0
3 years ago
Select all correct statements below:
Dafna11 [192]

Answer:

Explanation:

2. The image formed in a diverging lens is always virtual.

4. Converging lenses are shaped such that a beam of parallel light rays passing through the lens will be brought together in one single point.

8 0
4 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
3 years ago
"It is impossible to devise a process which may convert heat, extracted from a single
Vladimir [108]

Answer:

According to the second law of thermodynamics, we are unable to use the heat of the ocean and the atmosphere because we do not have a reservoir that has a temperature lower than the ocean or the atmosphere.

Explanation:

As you already know, the ocean and atmosphere have a lot of thermal energy, however, we are unable to convert this energy into mechanical energy that would be useful for our activities. This can be explained by the second law of thermodynamics, since it states that the presence of two bodies with different temperatures is necessary for it to be possible to transform heat into work.

In this case, to transform the thermal energy of the ocean and the atmosphere into mechanical energy we would need the existence of a thermal motor, which is only possible to be established when there is a body with high thermal energy and a sink, a reservoir, with low thermal energy, which will be the place where the heat will be expelled, to be converted into work. We do not have a reservoir with less thermal energy than the ocean and the atmosphere, so we cannot use their energy.

7 0
3 years ago
An 89.0 kg fullback moving east with a speed of 5.6 m/s is tackled by an 85.0 kg opponent running west at 2.84 m/s, and the coll
Korolek [52]

Answer:

a. v_f=1.477m/s

b. ΔK=1558.3J

c. E_k=1034.7 J

Explanation:

a).

Momentum conserved

p_{ix}=p_{fx}

m_1*v_1+m_2*v_2=(m_1+m_2)*v_f

v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}

v_f=\frac{89.0kg*5.6m/s+85.0kg*-2.84m/s}{(89.0+85.0)kg}

v_f=1.477m/s

b).

ΔK=K_i-K_f

\frac{1}{2}*m_1*v_1^2+\frac{1}{2}*m_2*v_2^2=\frac{1}{2}*(m_1+m_2)*v_f^2

\frac{1}{2}*89.0kg*(5.6m/s)^2+\frac{1}{2}*85.0kg*(2.84m/s)^2=\frac{1}{2}*(89.0+85.0)kg*(1.447m/s)^2

ΔK=1558.3J

c).

E_k=\frac{1}{2}*89kg*(5.8m/s)^2-\frac{1}{2}*(85+89)kg*(1.44m/s)^2

E_k=1034.7 J

d).

All of which has been lost as mechanical energy, and is now thermal energy warmer football players, noise a loud crunch for example.

8 0
3 years ago
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