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3 years ago

HELP ITS FOR SCIENCE

Physics

2 answers:

yuradex [85]3 years ago

7 0

For number 2:
Speed is related to kinetic energy because kinetic energy is the energy of motion.

Hope this helped!

Rashid [163]3 years ago

7 0

Answer:

A wrecking ball has the capability to destroy a building due to the greater mass and kinetic energy than the yo-yo.

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Question:

iogann1982 [59]

The correct answer is:
<span>C: in the protons and neutrons of an atom

In fact, the nuclear energy refers to the binding energy of the nucleons (protons and neutrons) of an atom. The protons and the neutrons are held together by the strong nuclear interaction, one of the four fundamental forces of nature, and the energy associated to this interaction is called nuclear energy.

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7 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

2 years ago

Please help on this one?

kupik [55]

Answer: A)30V. First find the current of the circuit. I=V/R(total resistance). So I=60/120=0.5. Now to find voltage drop in R3 use ohms law as given. V(of 3)=(0.5)(60)=30V

7 0

2 years ago

Light rays travel from one medium into another and refract away from the boundary. What changes about the light to cause this re

vaieri [72.5K]

Answer:

Explanation:

I think its a its speed increases.....

8 0

2 years ago

Read 2 more answers

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

Nesterboy [21]