What force pulls all objects with mass toward<br> one another?

A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p

Andrei [34K]

Answer:

13.4 x 10 raise to power -19 C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is E = 214 N/M

, The decrease in electrical potential energy is d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C

5 0

3 years ago

A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri

Alla [95]

Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

$q=1\times10^{-8}C$

Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

7 0

3 years ago

A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2

butalik [34]

Answer:

Number of turns in secondary will be 7

Explanation:

We have given primary voltage $V_p=120volt$

Number of turns in the primary is $N_p=3575$

Secondary voltage is given $V_s=235mV=0.235volt$

We have to find the number of turns in secondary

We know that $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

So $\frac{3575}{N_s}=\frac{120}{0.235}$

$N_s=6.60$

As the number of turns can not be in decimal so number of turns will be 7

6 0

4 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

4 years ago

A person is standing on a spring bathroom scale on the floor of an elevator which is moving up and slowing down at the rate of 2

Alinara [238K]

Answer:

923.44N

Explanation:

Obviously Fn cannot equal Fg since the elevator is moving up

Where Fn is the force read on the scale

Fg is gravitational force=mg=77.6 x 9.8=760.48N

So we use this equation:

Fnet = Fn - Fg

ma = Fn - mg

77.6 x 2.1 = Fn – 760.48

162.96N = Fn – 760.48N

923.44 N = Fn

The answer is 923.44 N

6 0

3 years ago

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What force pulls all objects with mass toward one another?