Ask question

Ask question

All categories

3 years ago

13

If two bodies have different forces but same acceleration and only one force is known how do I find the other one?

1 answer:

mihalych1998 [28]3 years ago

5 0

Answer:

Use F = m*a

Explanation:

If you are given the mass of the two different bodies, you could rearrange these equations, plug in the mass of the unknown body to find the unknown force

You might be interested in

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

kaheart [24]

You want to draw a free body diagram of the forces on the sled in the horizontal x-direction.

If you visualize the system in an x-y coordinate plane, the force along the x-direction is the angle it makes with the x-axis multiples by the force.

The angle made with the x-axis is cosine of the angle theta.

Please see picture attached.

5 0

3 years ago

Read 2 more answers

Linear expansivity?<br>

shusha [124]

Linear expansivity, area expansivity and volume or cubic expansivity are

7 0

2 years ago

An athlete needs to lose weight and decides to do it by “pumping iron.” (a) How many times must an 90.0 kg weight be lifted a di

Wittaler [7]

Answer:

230kg would be the best answer

Explanation:

7 0

3 years ago

You could use an analytical or triple beam balance to determine a ___ called ____

GaryK [48]

Answer:

a and b are the correct answers

Explanation:

7 0

3 years ago

Read 2 more answers

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of the earth, produ

aalyn [17]

Answer:

324795 C

252.637820565 N/C

$2.235844712\times 10^{-9}\ C/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

R = Radius of Mars = $3.4\times 10^6\ m$

A = Area = $4\pi R^2$

$\phi$ = Electric flux = $3.67\times 10^{16}\ Nm^2/C$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=3.67\times 10^{16}\times 8.85\times 10^{-12}\\\Rightarrow q=324795\ C$

The charge is 324795 C

Electric field is given by

$E=\dfrac{\phi}{A}\\\Rightarrow E=\dfrac{3.67\times 10^{16}}{4\pi (3.4\times 10^6)^2}\\\Rightarrow E=252.637820565\ N/C$

The electric field is 252.637820565 N/C

Surface charge density is given by

$\sigma=\dfrac{q}{4\pi R^2}\\\Rightarrow \sigma=\dfrac{324795}{4\pi (3.4\times 10^6)^2}\\\Rightarrow \sigma=2.235844712\times 10^{-9}\ C/m^2$

The surface charge density is $2.235844712\times 10^{-9}\ C/m^2$

6 0

3 years ago