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3 years ago

If the distance between two charged particles is doubled, the force between them changes by a factor of

Physics

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

If the distance between two charged particles is doubled, the force between them changes by a factor of

Explanation:

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What is another name for the heat-trapping gases

marissa [1.9K]

Greenhouse Gases, on relation to Earth's atmosphere.

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3 years ago

an experiment is set up to measure the effect of a gasoline additive on fuel consumption rate. what would be the control in this

Tamiku [17]

The control setup in this experiment would be one tank that does not contain any of the additives. Since the tanks with the gasoline additives would need to be compared with a tank that is not affected by the results of these additives.

4 0

3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

$P=-F\times v$

$F=\dfrac{-P}{v}$

$F=\dfrac{-146157\ W}{32\ m/s}$

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0

3 years ago

A skydiver jumps out of a plane flying at an altitude of 5,500 meters. How fast is the skydiver going when they get to the groun

andrey2020 [161]

Answer:

2,500 feet (760 meters)

Explannation: <em>At about 2,500 feet (760 meters), the skydiver throws out a pilot chute, and it deploys the parachute. Its used to control the fall rate.</em>

4 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago