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stepan [7]
2 years ago
15

How much work was done on a charge of 3.0 to move it through 12v electric potential difference

Physics
2 answers:
erik [133]2 years ago
8 0

Answer:

  • Hence, the work done in moving the charge is 45 J.

<h2>( Too lazy to translate it to Spanish T_T )</h2>
Ray Of Light [21]2 years ago
8 0

Answer:

45 J

Explanation:

Potential difference (V)= 15 V

Charge (Q)= 3C

We know that

The potential difference(p.d) between two point in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other point.

Potential difference (V) = work done (W)/Charge (Q)

V= W/Q

Putting the values in the above formula we get:

15 V = W / 3 C

W= 15 V × 3 C

W= 45 J

Hence, the work done in moving the charge is 45 J.

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Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.
Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0
3 years ago
What are the 2 types of ions.
Alina [70]

Answer:

Anions have more electrons than protons and so have a net negative charge. Cations have more protons than electrons and so have a net positive charge. Zwitterions are neutral and have both positive and negative charges at different locations throughout the molecule.

Explanation:

4 0
3 years ago
Why did the waco tornado happen?
Natali5045456 [20]
it was too much water inside the clouds and the tornado happened
6 0
3 years ago
If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

There are 2*pi radians in 1 revolution so:

RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

So you need about 30 RPM to keep people from falling out the bottom

7 0
4 years ago
A 50 kg person receives an absorbed dose of gamma radiation of 20 millirads. What is the total energy absorbed?.
Masteriza [31]

For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads,  the total energy absorbed is mathematically given as

E=0.1457J

<h3>What is the total energy absorbed?</h3>

Generally, the equation for the total energy absorbed  is mathematically given as

E=mass*gamma radiation

Therefore

E=50*20*19^{-3}

E=0.1457J

In conclusion,  the total energy absorbed

E=0.1457J

Read more about Energy

brainly.com/question/13439286

4 0
2 years ago
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