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2 years ago

How would you use one of Newton’s laws to build a fast car?

2 answers:

5 0

The second law when a force is applied to a car the change in motion is proportional to the force divided in the mass of the car as long as expressed by the famous equation F equals MA we’re F is for stem is the mass of the car and a is the acceleration a change in the motion of the car

kodGreya [7K]2 years ago

3 0

Answer:

The second law: When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car. This law is expressed by the famous equation F = ma, where F is a force, m is the mass of the car, and a is the acceleration, or change in motion, of the car.

Explanation:

Hope this helps

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A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1

nydimaria [60]

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

5 0

2 years ago

Does kinetic energy decrease as speed increase

leva [86]

Potential energy increases as speed decreases. Kinetic increases when speed increases.

6 0

2 years ago

A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

$Xc = \frac{1}{2\pi fC}$

where,

$Xc\\$ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

$Xc$ =?

$Xc' = \frac{1}{2\pi f'C}$

$= \frac{1}{2\pi 2f C}$

$= \frac{1}{2} (\frac{1}{2\pi fC} )\\$

$Xc' = \frac{1}{2} Xc$

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?