All categories

3 years ago

When one force acts, a second force is produced that is _______ and opposite to the original force?

Physics

2 answers:

Marianna [84]3 years ago

7 0

When one force acts, a second force is produced that is D.equal and opposite to the original force.

This is the answer to your question, have a great day, chao!!

lianna [129]3 years ago

6 0

Answer:

The answer is: D. Equal.

Explanation:

The force that generates the action produces a force of equal magnitude and opposite on the first.

The answer is: D. Equal.

You might be interested in

The chart below shows the amount of coal mined in four different states of USA in the year 2005.

ozzi

The correct answer is Montana.

In fact, we see from the table that Montana is the state with the largest amount of coal mined in 2005: 120 billion tons, while all the other states have a smaller amount of coal mined (37, 68 and 78, for West Virginia, Wyoming and Illinois, respectively).

3 0

3 years ago

Read 2 more answers

Which is a vector quantity, speed or velocity? defend your answer.

krek1111 [17]

Speed is the distance traveled divided by the time it took to travel this distance. Velocity is the change in position divided by the time of travel. Velocity only depends on the starting and ending point but the speed depends on the path. Speed is a scalar quantity (the distance per time ratio) and velocity is a vector quantuty, because it is defined also with its direction.

4 0

3 years ago

You are given four resistors, 2 ohms, 3 ohms, 5 ohms, and 10 ohms. Your friend say you can connect them so you obtain an equival

AlexFokin [52]

If we will connect the resistors 2ohms, 3ohms, 5ohms in series and the 10ohms resistance parallel then we get equivalent resistance of 5 ohms.

The equivalent circuit is,

R equivalent for the series connection is,

$\begin{gathered} Req(S)=2+3+5=10ohms \\ Now,\text{ } \\ Req\text{ }for\text{ }10\text{ }ohms\text{ }and\text{ }10\text{ }ohms\text{ }is, \\ \frac{1}{Req}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ So, \\ Req=5\text{ ohms} \end{gathered}$

The equivalent resistance is 5 ohms.

So your friend is saying true.

6 0

10 months ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

ValentinkaMS [17]

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

<h3>Acceleration of the box</h3>

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

<h3>Time of motion of the box</h3>

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

<h3>Average power supplied by the friction</h3>

P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

Learn more about average power here: brainly.com/question/19415290

#SPJ1

7 0

1 year ago

||||| A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless

amid [387]

Answer

The lighter block will have the Positive Acceleration that is +2.45 m/s square.

A (1.5 Kg) = + 2.45 m/s square

Explanation:

To solve this there is a hard way to do this and there is an easy way to do this. The hard way is to solve Newton's second law for each block individually and then combine them and you get two equations with two unknowns. you try your best to solve the algebra without losing any sins but lets be honest it usually goes wrong.

So the easy way to do this the way to get the magnitude of the acceleration of the blocks. That is to say that i want to know the magnitude at which 2.5 Kg block accelerates or 1.5 Kg block accelerates when the the blocks were released.

Take the net external force that tries to make system go and divide it by total mass of the the system.

A (of System) = F(net external force) / m (total mass of system)

This is the quick way to know the magnitude of acceleration of the objects in the system. But this is only possible if the system is moved in same magnitude of acceleration that is 2.5 Kg block will move downward and 1.5 Kg block will move upward with the same magnitude. So here in this case we have friction-less pulley and the blocks will move with the same magnitude of acceleration.

To find the external forces

External forces are the forces which exerted on the objects in our system from the objects outside of our system. So one external force is the force of gravity. Both 2.5 Kg block and the force of gravity will be in downward direction.

Force of gravity on 2.5 Kg block

F = + (2.5 x 9.8) = 24.5

After releasing the rope the 2.5 Kg block will drive the system and accelerates in downward direction so It will be a positive force.

Force of gravity on 1.5 Kg block

F = - (1.5 x 9.8) = 14.7

The force of gravity on 1.5 Kg block will be negative because it will accelerate in upward or opposite direction of the force of gravity. Because the whole system is moving in one direction but the force of gravity on 1.5 Kg block is opposing the acceleration of the system.

Now divide the the Net external forces by total mass of the blocks that is

A (of System) = F(net external force) / m (total mass of system)

A = (+ 24.5 - 14.7) / 2.5 + 1.5

A = 9.8 / 4

A = 2.45 meter per second square

So the Acceleration of 2.5 Kg block will be Negative that is -2.45 m/s square. Since block is accelerating down and we usually treat down as negative.

A (1.5 Kg) = - 2.45 m/s square

So the Acceleration of 1.5 Kg block will be Positive that is +2.45 m/s square. Since block is accelerating up and we usually treat up as positive.

A (1.5 Kg) = + 2.45 m/s square

4 0

3 years ago