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tatyana61 [14]
3 years ago
11

You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. What

is the direction of the force on the charge you are holding?A. to the eastB. to the southwestC. to the southD. to the northeast
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer: Option (B) is the correct answer.

Explanation:

It is known that Colombian force between similar charges is repulsive in nature. Therefore, it means that the direction of force experienced by the charges in North and East will be South and West respectively.

As a result, this means that the forces are going to be towards the south-west direction.

Thus, we can conclude that if you are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. The direction of the force on the charge you are holding should be in south-west direction.

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Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at
klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

\lambda = Wavelength

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

Energy of a photon is given by

E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}

Let \lambda_1 = 700 nm

\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}

For red light

E_1=\dfrac{hc}{\lambda_1}

For UV light

E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}

Dividing the equations

\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1

Hence, the answer is  b) twice the energy of each photon of the red light.

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3 years ago
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You weigh 580 N on Earth. If you were to go to Mars, where its gravitational pull is 3 . 7 11 m /s 2 , what would you weigh? (Hi
andrew-mc [135]

Answer:

59.18 kg

Explanation:

use f=ma

f= 580 N

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

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2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

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Answer:

A

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B. Some of the ball’s energy is transformed to thermal energy.

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