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dimulka [17.4K]

2 years ago

8

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

n zero.equal to zero.c. equal to the average of the two distances times the charges.d. equal to the net electric field.

1 answer:

Marizza181 [45]2 years ago

8 0

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

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A baseball with a mass of 0.80 kg is given an acceleration of 20.00 m/s. How much net force was applied to the ball

dybincka [34]

a x m = f

.80 x 20 = 16

4 0

2 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

2 years ago

HElP I NEED TO turn this in

Alexeev081 [22]

B student 2 because you add

8 0

2 years ago

A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for

Norma-Jean [14]

Answer:

Explanation:

Let's analyze the situation presented in order to know which answer is correct.

When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.

Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.

The correct answer is:

C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck

4 0

2 years ago

Looking at the position of the house and the tree, if the man ran starting from the house going

Aleks04 [339]

Average velocity of man is 2 m/s

<u>Given that;</u>

Distance between tree and house = 16 m

Time taken = 8 seconds

<u>Find:</u>

Average<em> velocity</em>

<u>Computation:</u>

Average velocity = Total distance cover / Time taken

Average velocity = Distance between tree and house / Time taken

Average velocity = 16 / 8

Average velocity = 2 m/s

Learn more:

brainly.com/question/22610586?referrer=searchResults

4 0

1 year ago