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OverLord2011 [107]
2 years ago
15

10 C of charge are placed on a spherical conducting shell. A point particle with a charge of –3C is placed at the center of the

cavity. The net charges, in coulombs, on the inner and outer surfaces of the shell, is:
a. -7
b. +7
c. 0
d. +13
e. +6.5
Physics
1 answer:
fenix001 [56]2 years ago
6 0

Answer:

The net charges on inner surface is +3 C

The net charges on outer surface is +7 C

Explanation:

Given that,

Charge on spherical conducting shell = 10 C

Charge at center = -3 C

We know that,

The net electric field inside the conducting shell is zero. The Gaussian surface inside the conductor must have zero net charge.

We need to calculate the net charges on inner surface

The charge on the inner surface of the conductor must be equal and opposite to the present charge of the  center.

So, The net charges on inner surface is +3 C.

We need to calculate the net charges on outer surface

Using formula of net charge

Q'=Q+q

Put the value into the formula

Q'=10-3

Q'=+7C

Hence, The net charges on inner surface is +3 C

The net charges on outer surface is +7 C

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Ben starts walking along a path at 3 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the s
Drupady [299]

Answer:

3 hours

Explanation:

Given:

- The speed of Ben v_b = 3 mi/h

- The speed of Amanda v_a = 6 mi/h

- The total time taken to cover distance(d) by ben = t_b

Find:

How long will it be before Amanda catches up to Ben?

Solution:

- The distance d traveled by Ben:

                                 d = v_b*t_b

                                 d = 3*t_b

- The distance d traveled by Amanda:

                                 d = v_a*t_a

                                 d = 6*t_a

- Equate the distance as when they meet:

                                 3*t_b = 6*t_a

- Where ,

                                  t_b = t_a + 1.5

                                  t_a = t_b - 1.5

- Substitute the time relationship in distance relationship:

                                  3*t_b = 6*(t_b - 1.5)

                                  3*t_b = 6*1.5

                                      t_b = 2*1.5 = 3 h

- Hence, It would take 3 hours since Ben starts walking that amanda catches up.

4 0
2 years ago
A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow
valentina_108 [34]
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
7 0
2 years ago
Read 2 more answers
"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression
zavuch27 [327]

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

6 0
2 years ago
jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en
d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

\theta = \frac{2\pi}{6}

\theta = \frac{\pi}{3}

so the angular speed of the train is given as

\omega = \frac{\theta}{t}

\omega = \frac{\pi}{30} rad/s

now we have time period of the train given as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{\frac{\pi}{30}}

T = 60 s

5 0
2 years ago
A stranded soldier shoots a signal flare into the air to attract the attention of a nearby plane. The flare has an initial verti
Ipatiy [6.2K]

Answer:

Explanation:

h = ut - 16 t² = ut - 1/2 x32 t² = ut - 1/2 g t² , g = acceleration  = - 32 ft / s²

1) v² = u² - 2 g h , v = 0 so

h = u² / 2g = 1500² / 2 x 32 = 35156.25 ft

2) v = u - gt

t = u / g = 1500 / 32 = 46.875 s

3) It will hit the ground after 2 x 46.875 = 93.75 s

4 ) time to reach 30000 ft height  t is given by

h = ut - 16 t²

30000 = 1500t - 16t²

16t²-1500t + 30000 = 0

t = 28.92 s  and 64.82 s

Time required to travel 50000 by plane

= 50000/880 = 56.82 . There is no match of timing so plane will not hit it.

8 0
2 years ago
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