All categories

3 years ago

How much work is required to move a 1 nC charge from an electric potential of 0 V to a potential of 18V?

Physics

1 answer:

jolli1 [7]3 years ago

3 0

ni work is required by an external force because this is the natural movement of the charge

You might be interested in

A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 190m. Find the the acceleration of t

Mama L [17]

Answer:

a = 7.75 [m/²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}$

where:

x = final distance = 190 [m]

Xo = initial distance = 0

Vo = initial velocity = 0 (car starts from the rest)

a = acceleration [m/s²]

t = time = 7 [s]

190 = 0 + (0*7) + 0.5*a*(7²)

190 = 0.5*49*a

a = 7.75 [m/²]

4 0

3 years ago

A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6

steposvetlana [31]

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)

T = 24.9 N

6 0

3 years ago

Two large parallel metal plates carry opposite charges. They are separated by 0.20 m and the p.d. between them is 500 V. What is

Bezzdna [24]

Hi there!

We can use the following relationship between the Potential Difference and the Electric field:

$V = E d$

V = Potential Difference (500V)
E = Electric Field (V/m)
d = separation between plates (0.2 m)

We can rearrange the equation to solve for the electric field:
$E = \frac{V}{d}\\\\$

Plug in the given values.

$E = \frac{500}{0.2} = \boxed{2500 \frac{V}{m}}$

3 0

2 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

MArishka [77]

Answer:

A) F_g = 8.9278 × 10^(-30)

B) F_e = 1.6 × 10^(-17) N upwards

C) F_m = 6.8 × 10^(-17) N downwards

Explanation:

A) Formula for gravitational force is;

F_g = m_e × g

m_e is mass of electron = 9.11 x 10^(-31) kg

F_g = 9.11 x 10^(-31) × 9.8

F_g = 8.9278 × 10^(-30) N downwards

B) Formula for Electric force is;

F_e = qe

q is charge on electron = 1.6 × 10^(-19) C

E is electric field = 100 N/C

F_e = 1.6 × 10^(-19) × 100

F_e = 1.6 × 10^(-17) N upwards

C) Magnetic force is given by the formula;

F_m = qVB

q is charge on electron = 1.6 × 10^(-19) C

V is velocity given as 8.50 × 10^(6) m/s

B is magnetic field = 50.0 μT = 50 × 10^(-6) T

F_m = 1.6 × 10^(-19) × 8.50 × 10^(6) × 50 × 10^(-6)

F_m = 6.8 × 10^(-17) N downwards

6 0

3 years ago

A GPS receiver can tell you your exact latitude and longitude?

Dmitry_Shevchenko [17]

That is false. A GPS cant tell latitude and longitude

3 0

3 years ago