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Leni [432]
3 years ago
6

What is the equation for measuring the change in thermal energy?

Physics
1 answer:
Daniel [21]3 years ago
7 0

Answer:

Explanation: I think...

Thermal Energy formula Q = mcΔT

Q = Thermal Energy(J)

m = Mass(kg)

c = Specific Heat(J/kg°C)

ΔT = Change in Temperature(°C)

you have to write the equation based on what you are working on

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Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a second identical ball (Ball B) initially at rest. Afte
timurjin [86]

Answer:

Explanation:

We shall apply law of conservation of momentum during the collision of ball A and B .

Total momentum before collision of A and B = .35 x 10 = 3.5 kg m/s

Let the velocity of B after collision be v .

Total momentum after collision  = .35 x 2 + .35v

According to law of conservation of momentum

.35 x 2 + .35v  = 3.5

.35 v = 2.8

v = 8 m /s .

The direction of B will be same as direction of A .

3 0
3 years ago
Which of the following is a homogenous mixture?
Ivanshal [37]

Answer:

Blood is a homogenous mixture

3 0
3 years ago
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James is planning a science fair project on sound waves. He places an alarm inside a jar which he can remove the
otez555 [7]

Answer: c

Explanation:

Sound waves cannot travel through a medium

3 0
3 years ago
This is a question on my physics test :)
Licemer1 [7]

Answer:

119.6 J/Kg°C

Explanation:

Data obtained from the question include:

Mass of substance (ms) = 170 g

Initial temperature of substance (Ts) = 120 °C

Volume of water = 200 mL

Initial temperature of water (Ts) = 10 °C

Temperature of the mixture (T2) = 12.6 °C

Density of water = 1 g/mL

Specific heat capacity of water (Cw) = 4200J/Kg°C

Specific heat capacity of substance (Cs) =..?

Next, we shall determine the mass of water. This can be obtained as follow:

Volume of water = 200 mL

Density of water = 1 g/mL

Mass of water =..?

Density = mass /volume

1 = mass /200

Cross multiply

Mass of water = 1 x 200

Mass of water = 200 g

Convert 200 g of water to Kg

Mass of water = 200/1000 0.2 Kg

Mass of water = 0.2 Kg

Now, we obtained the specific heat capacity of the substance using the following formula:

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

Mass of water = 0.2 Kg

Initial temperature of water (Ts) = 10 °C

Specific heat capacity of water (Cw) = 4200J/Kg°C

Temperature of the mixture (T2) = 12.6 °C

Mass of substance (ms) = 170 g = 170/1000 = 0.17 Kg

Initial temperature of substance (Ts) = 120 °C

Specific heat capacity of substance (Cs) =..?

MwCw(T2 – Tw) + MsCs(T2 – Ts) = 0

0.2× 4200(12.6 – 10) + 0.17×Cs×(12.6 – 120) = 0

840(2.6) + 0.17Cs(– 107.4) = 0

2184 – 18.258Cs = 0

Rearrange

2184 = 18.258Cs

Divide both side by the coefficient of Cs i.e 18258

Cs = 2184/18.258

Cs = 119.6 J/Kg°C

Therefore, the specific heat capacity of the substance is 119.6 J/Kg°C

7 0
3 years ago
Two protons move at speed v, one along the x axis and one along the y axis. What is the relative speed of the two protons? Evalu
forsale [732]

Answer:

0.85 c

Explanation:

When the two travel at right angles to each other , their relative velocity is given by the hypotenuse formed by the x and y direction velocities.

Relative velocity of the two protons is  given as

\sqrt{v^{2}+v^{2}} = (\sqrt{2} ) v

If v = 3/5 c , then the relative speed = \sqrt{2} (3/5c) =0.85 c

8 0
3 years ago
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