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vova2212 [387]
3 years ago
8

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

Physics
2 answers:
solmaris [256]3 years ago
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]3 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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Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut
igomit [66]

Answer:

Muatan listrik yang mengalir dari potensial tinggi ke potensial rendah disebut

elektron

Explanation:

semoga ini membantu (:

8 0
3 years ago
2 kids are running down the hall at 2 m/s, one is 40kg and the other is 50kg (p=ma)
Georgia [21]

The net force acting on the two (2) kids is equal to 180 Newton.

<u>Given the following data:</u>

  • Mass of kid 1 = 40 kilogram.
  • Mass of kid 2 = 50 kilogram.
  • Acceleration = 2 m/s^2

To determine the net force acting on the two (2) kids:

First of all, we would calculate the total mass of the two (2) kids:

Total\;mass = 50+40

Total mass = 90 kg.

<h3>How to calculate net force.</h3>

In this exercise, you're required to calculate the magnitude of the net force that is acting on these two (2) kids running down the hall. Thus, we would apply Newton's Second Law of Motion.

Mathematically, Newton's Second Law of Motion is given by this formula;

Net\;force = mass \times acceleration

Substituting the given parameters into the formula, we have;

Net\;force = 90 \times 2

Net force = 180 Newton.

Read more on acceleration here: brainly.com/question/1121817

7 0
3 years ago
If the value of the resistor r2 were doubled, how would the value of the resistor r3 have to change in order to keep the current
Reika [66]

This is a Wheatstone bridge, and the ratio of R2 to R1 equals the ratio of Rx to R3. As a result, if R2 is increased, R3 should be reduced by a factor of two.

<h3>Explain Wheatstone bridge?</h3>

A Wheatstone bridge is a type of electrical circuit that is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one of which contains the unknown component.

The Wheatstone bridge circuit can be used to compare an unknown resistance RX to others of known value, such as R1 and R2, which have constant values and R3 which can be variable.

If we connected a voltmeter, ammeter, or galvanometer between points C and D, and then changed resistor R3 until the meters read zero, the two arms would be balanced, and the value of RX (substituting R4) would be known as indicated.

To learn more about Wheatstone bridge refer to :

brainly.com/question/15225070

#SPJ4

5 0
1 year ago
A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20, what is the exact distance that the seismic station is away from
maxonik [38]

The exact distance from the seismic station to the epicenter is 6000 km.

<h3>Epicenter of earthquake</h3>

The earthquake's epicenter is the point above the fault location on the earth's surface.

Given that:

A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20.

Difference in arrival time = 3:07:20 - 3:00:00 = 7 minutes 20 seconds

From the earthquake time travel graph, a time difference of 7 min 20 sec is at x = 6

The exact distance from the seismic station to the epicenter is 6000 km.

Find out more on Epicenter of earthquake at: brainly.com/question/1969968

4 0
2 years ago
Describe how air pressure is exerted on objects.
Delicious77 [7]
<span>The same amount of pressure is exerted on all sides of the object equally. If you had 13 psi, then that means on the top of the object, 13 psi would be pushing in on the object. From the bottom, 13 psi would be pushing up on the object. Same goes for all the sides as well. The pressure pushes 'in' on the object.</span>
5 0
3 years ago
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