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MakcuM [25]
3 years ago
7

Thermal effects refers to the:

Physics
2 answers:
a_sh-v [17]3 years ago
4 0

Answer: C.removal of heat by cooling towers

The thermal effect refers to the thermal radiations emitted by the body. The thermal effect refers to the heat lost by the body or object. The removal of heat by cooling towers is an example of thermal effect as cooling tower is a body from which heat is being lost or removed due to thermal effect.

Marat540 [252]3 years ago
3 0

The correct answer is actually C. Not A.

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4 0
3 years ago
Will give correct answer brainliest​
lukranit [14]

Answer:

I THINK it’s A

Explanation:

Because all the other answers don’t make sense.

5 0
3 years ago
If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?
Bess [88]

This is an incomplete question, here is a complete question.

The air in the cylinder with a piston has a volume of 220 mL and a pressure of 650 mmHg.

If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?

Answer : The final volume of the cylinder is, 117.6 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas = 650 mmHg = 0.855 atm   (1 atm = 760 mmHg)

P_2 = final pressure of gas = 1.6 atm

V_1 = initial volume of gas = 220 mL

V_2 = final volume of gas = ?

Now put all the given values in the above equation, we get:

0.855atm\times 220mL=1.6atm\times V_2

V_2=117.6mL

Therefore, the final volume of the cylinder is, 117.6 mL

8 0
3 years ago
Andrea was watching her brother in the ocean and noticed that the waves were coming into the beach at a frequency of 0.777778 Hz
Stolb23 [73]
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3 0
3 years ago
A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at
wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 =  \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s -  2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0
3 years ago
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