Ask question

Ask question

All categories

3 years ago

13

If the work required to move a 0.35 c charge from point a to point b is 125 j, what is the potential difference between the two

points?

1 answer:

valentina_108 [34]3 years ago

7 0

Using W = qV => V = W/q

You might be interested in

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.

sergeinik [125]

Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²

5 0

3 years ago

If y^2= 3.249 x 10^-11, y = ?

Leni [432]

Answer:

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

5 0

3 years ago

Read 2 more answers

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?

Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

$W=F\times x$

$W=mg\times x$

Where, x = distance

mg = weight

Put the value into the formula

$W=40\times2$

$W= 80\ Nm$

$W=80\ J$

Hence, The work done on the box is 80 J.

5 0

3 years ago

Which describes electricity?

SVETLANKA909090 [29]

Answer:

all of the above

Explanation:

- a build up of electric charge

- the force and motion of electrically charged particles

- an electric current

are three different ways to describe electricity.

So the answer is all of the above.

7 0

2 years ago