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Ipatiy [6.2K]
2 years ago
11

3 help with physics!! ---------------- will give brainliest

Physics
1 answer:
Zinaida [17]2 years ago
6 0

Answer:

Answer B

Explanation:

An increase in resistance makes it harder for the electric current to pass through

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What is the acceleration of a car that starts from rest and attains a final speed of 20 m
OlgaM077 [116]

Answer:

323.9

Explanation:

3 0
3 years ago
An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e
Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

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t = 15.2 s

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Which of the following pulley systems have/has a greater mechanical advantage than the one shown above?
astraxan [27]
Is there a picture or description?

8 0
3 years ago
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A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease
skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm

So, the new pressure is 2 atm.

7 0
2 years ago
Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
2 years ago
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