3 years ago

How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?

Physics

1 answer:

rjkz [21]3 years ago

4 0

Answer:

6.3E-6

Explanation:

Workdone = V Q

You might be interested in

What is the science behind the making of pop-up books?

IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0

3 years ago

A ball on an extremely low-friction, tilted surface will slide downhill without rotating. If the surface is rough, however, the

spayn [35]

It will go left or right

6 0

3 years ago

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumpe

Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

$U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where we have

$k=2\cdot 10^6 N/m$ is the spring constant of the bumper

x is the maximum compression of the bumper

$m=30\cdot 10^4 kg$ is the mass of the car

$v=2.0 m/s$ is the speed of the car

Solving for x, we find the maximum compression of the spring:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m$

8 0

3 years ago

Read 2 more answers

The following equation, N2 + 3 H2 —>2 NH3 ,describes a

mafiozo [28]

Physical change 1 is the answer

4 0

2 years ago

Read 2 more answers

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

3 0

3 years ago