Ask question

Ask question

All categories

Natali5045456 [20]

3 years ago

9

During a basketball game what are the most points you can score from shooting and getting fouled

1 answer:

Illusion [34]3 years ago

8 0

Answer:

If a player is fouled while shooting a three-point shot and makes it anyway, he is awarded one free throw. Thus, he could score four points on the play. Inbounds- If fouled while not shooting, the ball is given to the team the foul was committed upon.

You might be interested in

What does it mean that are bodies are in osmotic equilibrium but chemical and electrical disequilibrium? Why? What molecules are

Whitepunk [10]

Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

8 0

3 years ago

Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST

Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

$a^2+b^2=c^2$

Plug values in

$88.4^2+174^2=c^2$

Simplify

$7814.56+30276=c^2$

Add the two values

$38090.56=c^2$

Take the square root of both sides

$195.168\approx195.168$

8 0

2 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

What are the conditions to increase the stability of a body?

cricket20 [7]

Training everyday
..........

5 0

3 years ago

A 2kg book is moved from a shelf that is 2m off the ground to a shelf that is 1.5m off the ground, what is it’s change in gravit

Ket [755]

The gravitational energy is going up subtracting the energy that was on the ground

3 0

4 years ago