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3 years ago

What is the direction of friction if a ball is rolling on the ground?

1 answer:

3 0

The direction is the same as the direction the ball is moving in. Because of the rolling of the ball, the direction of movement of the surface of the ball is opposite the overall direction of the ball. Since friction will oppose the direction of movement of the surface of the ball, it is in the same direction as the net direction of movement of the ball.

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A storm is moving at 15 km/hr what to do to determine its velocity

WINSTONCH [101]

Check the current weather map for 2 different times, and see where the center of the storm is. That tells you what direction it's moving. With its speed and direction, you have its velocity.

6 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

In which regions can the gravitational field strength due to the two planets be zero? Explain.

yan [13]

Answer:

I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong

Explanation:

7 0

3 years ago

Increasing the concentration of greenhouse gases in Earth's atmosphere decreases the transparency of the atmosphere to infrared

kondaur [170]

Answer:

Decreases the transparency of the atmosphere to infrared light.

Explanation:

When a large amount of green-house gases are present in the atmosphere, the layer of these gases become opaque to infrared radiation and radiation from the sun get trapped into these gases molecules. These excited molecules radiate this energy into our own atmosphere and that why the temperature of Earth is rising due to the Green-House effect.

8 0

3 years ago

A book is sitting on a desk. What best describes the normal force acting on the book?

hjlf

Answer:

umm im not sure

3 0

3 years ago