3 years ago

What distance does a biker travel if he rides at a constant speed or 22 m/s for 45 seconds?

Physics

1 answer:

vredina [299]3 years ago

5 0

Answer:

it would be 990 m.

Explanation:

22 m/s x 45 seconds.

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A 1.5m wire carries a 4 A current when a potential difference of 66 V is applied. What is the resistance of the wire?

kifflom [539]

Resistance = Voltage / current
Resistance = 66/4
=16.5ohms
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3 years ago

If Joey ran 60 meters in 10 seconds, then his velocity is ________ m/s.

BabaBlast [244]

Answer:

6 m/s

Explanation:

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Which of the following is not a constant or control for this experiment?

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D.the amount of time in sun

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A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

madreJ [45]

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

5 0

3 years ago

A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w

user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

$p\propto T$

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where in this problem:

$p_1=0.981 atm$ is the initial pressure of the gas

$T_1=65^{\circ}+273=338 K$ is the initial absolute temperature of the gas

$T_2=41^{\circ}+273=314 K$ is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

$p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm$

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3 years ago