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Advocard [28]
3 years ago
7

In order to increase thermal energy, the temperature must ___________.

Physics
2 answers:
Pie3 years ago
8 0

Answer: Option (a) is the correct answer.

Explanation:

Energy obtained by an object due to the motion of particles that are present within the object is known as thermal energy.

So, when we increase the temperature of a substance then its molecules tend to gain kinetic energy. Because of this molecules will start to move from their initial position and hence, there will be more number of collisions between the molecules.

As a result, there will be increase in thermal energy of the substance.

Thus, we can conclude that in order to increase thermal energy, the temperature must increase.

Anni [7]3 years ago
5 0
In order to increase thermal energy, the temperature must <span>increase</span>
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Which of these forms of radiation passes most easily through the disk of the Milky Way?
Rama09 [41]
<h2>Answer: 3 - infrared light</h2>

Explanation:

<u>There are certain areas of the Milky Way that cannot be observed using the visible range of the electromagnetic spectrum</u> (this includes blue light and red light). This is because these areas are covered or hidden behind columns of interstellar dust and dark matter.

However, using infrared light and sometimes radio waves, it is possible to observe the galaxy better, because this light manages to pass through all that interstellar dust.

4 0
3 years ago
A machine has mechanical advantage 2.What does that mean​
EleoNora [17]

Answer:

Mechanical advantage is the measure of the force amplification achieved by using a tool , mechanical device or machine . The machine preserve the input power and supply trade off force. against movement to obtain a desired amplification in the output force .

5 0
3 years ago
Read 2 more answers
A particle P with speed 140 m s–1begins to decelerate uniformly at a certain instant while another particle Q starts from rest 6
Natasha2012 [34]

Answer:

i) The motion of both particles are shown on the same speed-time curve included

ii) Approximately 19.5 seconds

Explanation:

We are given that;

Initial velocity of particle, P = 140 m/s

Start time of particle P = 6 s before start time of particle Q

Position of particle Q when velocity is 25 m/s = 125 m

Therefore, from the equation of motion, we have for particle Q;

v² = u² + 2·a·s

Where:

v = Final velocity = 25 m/s

u = Initial velocity = 0 m/s

a = Acceleration

s = Distance covered = 125 m

Therefore;

25² = 0² + 2×a×125

Which gives a = 25²/(2×125) = 2.5 m/s²

The time taken for particle Q to reach 125 m is found from the relation;

s = u·t + 1/2·a·t²

Where:

t = Time of journey

Therefore;

125 = 0×t + 1/2×2.5×t²

Which gives 125 = 1.25 × t²

Hence, t² = 125/1.25 = 100

t = √(100) = 10 s

The equation for particle Q is v = 0 + 2.5×t

Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;

Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s

From the equation of motion, for particle P (decelerating) we have

v = u - a·t

Where:

v = 25 m/s

u = 140 m/s

t = 16 s

Hence, 25 = 140 - a×16

∴ 16·a = 140 - 25 = 115

a = 115/16 = 7.1875 m/s²

Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;

v = u - a·t

0 = 140 - 7.1875 × t

∴7.1875·t = 140

t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.

4 0
3 years ago
A jar of tea is placed in sunlight until it
Tanzania [10]

Answer:

m = 4.44 [g]

Explanation:

This is a problem related to heat transfer and thermodynamics. The second law of thermodynamics tells us that heat goes in only one direction, from the highest temperature substance to the lowest temperature. In this case the heat goes from tea to ice. The heat transfer process could be defined as the heat rejection of one body will be equal to the Heat received by another body.

Q_{out}=Q_{in}\\Q_{out}=m*cp*(T_{f}-T_{i})

Where:

m = mass = 177[g] = 0.177[kg]

Cp = specific heat = 4186 [J/kg*C]

T_f = 29.1 [°C]

Ti = 31.1 [°C]

Q = 0.177*4186*(29.1 - 31.1)

Q = - 1481.8 [J]

Note: The negative sign means that the heat is rejected.

Recall that the heat rejected is equal to the heat transferred

As the ice is going through a phase change, the fusion latent heat should be used, i.e. when it passes from solid to liquid. The heat transfer for this process is calculated with the following expression.

Qin = m *hf

m = mass of the ice [kg]

hf = Fusion latent heat = 333700 [J/kg]

1481.8 = m * 333700

m = 0.00444 [kg] = 4.44 [g]

7 0
3 years ago
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m
Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

       = 4.98 × 10¹⁹ protons

       ≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

                                                         = 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

                                                         = 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

     = 30 × 10⁷/14.62

     = 2.05 × 10⁷ °C

5 0
3 years ago
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