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IgorLugansk [536]
3 years ago
5

If the same net force acts on an object, what happens to its acceleration as its mass increases?

Physics
2 answers:
strojnjashka [21]3 years ago
6 0

Answer:

The answer is B- Decreases

ANEK [815]3 years ago
3 0
The answer is B- Decreases
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If the ac peak voltage across a 100-ohm resistor is 120 V, then the average power dissipated by the resistor is ________
Kazeer [188]

Answer:

The average power dissipated is 72 W.

Explanation:

Given;

peak voltage of the AC circuit, V₀ = 120 V

resistance of the resistor, R = 100 -ohm

The average power dissipated by the resistor is given by;

P_{avg} = \frac{1}{2} I_oV_o= I_{rms}V_{rms} = \frac{V_{rms}^2}{R}

where;

V_{rms} is the root-mean-square-voltage

V_{rms} = \frac{V_o}{\sqrt{2}} \\\\V_{rms} = \frac{120}{\sqrt{2}}\\\\V_{rms} = 84.853 \ V

The average power dissipated by the resistor is calculated as;

P_{avg} = \frac{V_{rms}^2}{R}\\\\P_{avg} = \frac{84.853^2}{100}\\\\P_{avg} = 72 \ W

Therefore, the average power dissipated is 72 W.

5 0
3 years ago
Sum of the length of all three sides of the of a triangle is known as​
Klio2033 [76]

Answer:

It's perimeter

Explanation:

That's what it said when I looked it up so hopefully that's correct.

3 0
3 years ago
Which type of energy increases when an object's atoms move faster?
Travka [436]

Answer:

thermal

Explanation:

when an object's atoms move it causes friction, when friction happens, it heats ups causing thermal energy

8 0
3 years ago
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A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before co
Tema [17]
<span>This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a -- > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.</span>
3 0
3 years ago
A 2.0 kg mass attached to an ideal spring oscillates horizontally with an amplitude of 0.15 m. The spring constant is 85N/M
dimulka [17.4K]

Answer:

I think the answer is electric potential.

4 0
3 years ago
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