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mihalych1998 [28]
3 years ago
15

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

c force. What is the charge on the ball bearing?
Physics
1 answer:
Alla [95]3 years ago
7 0

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

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A sample container of carbon monoxide occupies a volume of 435 ml at a pressure of 785 torr and a temperature of 298 k. What wou
strojnjashka [21]

Answer:

181.54 K

Explanation:

From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then

T2=v2t1/ v1

Given information

V1 435 ml

V2 265 ml

T1 298K

Substituting the given values then

T2=265*298/435=181.54 K

3 0
3 years ago
A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express
andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

      E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

     E = mcΔT

     725 = 0.055 x 921.096 x (T - 27.5)

     T - 27.5 = 14.31

     T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

6 0
3 years ago
A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
Read 2 more answers
An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?
anyanavicka [17]

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

8 0
3 years ago
Read 2 more answers
In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion
Leno4ka [110]

Answer:

4 m/s²

Explanation:

 When the elevator is 1 m below point of contact , compression will be 1 m.

Restoring force in the spring will be 10600 N. Friction force of 17000N  will also act in upward direction . The weight of 2000 x 9.8 N will act downwards

Force in down ward direction = 2000 x 9.8

= 19600 N

Force in upward direction

= 10600 + 17000

= 27600 N

Net force in upward direction

= 27600 - 19600

= 8000 N

Acceleration in upward direction

= 8000 / 2000

= 4 m/s²

5 0
3 years ago
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