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arlik [135]
4 years ago
8

PLEASE help me with these questions as soon as possible i need to turn it in!!!!

Physics
1 answer:
Sloan [31]4 years ago
3 0
A. I believe this might be the answer but
6 J/60 J= .1
.1 x 100= 10%

b. I'm not sure of this question


19. I'm not sure if I did this correct way or if there is a better way (I'm rusty) 
300 J/.50 %= 600
600/100= 6 (answer)


20. (200 J)(.75%) = 150 (answer)
Check:
Efficiency: output/input x 100
150 J/200 J= .75
.75 x 100= 75%
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What is the resistance at 20°C of a 2.0-meter length of tungsten wire with a cross-sectional area of 7.9 10^-7
Bad White [126]

Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

8 0
3 years ago
What makes an electric charge?
scZoUnD [109]

Answer:

Most electric charge is carried by the electrons and protons within an atom. Conversely, two protons repel each other, as do two electrons. Advertisement. Protons and electrons create electric fields, which exert a force called the Coulomb force, which radiates outward in all directions.

5 0
3 years ago
Read 2 more answers
Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on
lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

4 0
4 years ago
If the phase of the vibrating sources was changed so that they were vibrating completely out of phase, what effect would this ha
Over [174]

Answer:

There would be complete destructive interference.

Explanation:

This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.

Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.

Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.

4 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
4 years ago
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