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balu736 [363]
4 years ago
14

A person is placed in a large, hollow, metallic sphere that is insulated from ground. (a) If a large charge is placed on the sph

ere, will the person be harmed upon touching the inside of the sphere? Yes No Correct: Your answer is correct. (b) Explain what will happen if the person also has an initial charge whose sign is opposite that of the charge on the sphere.
Physics
1 answer:
stiv31 [10]4 years ago
7 0

Answer:

Explanation:

It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.

If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.

If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.

If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere                        

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Your own car has a mass of 2000 kg. If your car produces a force of 5000 N, how fast will it accelerate?
Sidana [21]

Answer:

2.5m/s²

Explanation:

Given parameters:

Mass of car  = 2000kg

Force produced by the car  = 5000N

Unknown:

Acceleration of the car  = ?

Solution:

According to Newton's second law of motion, Force is a product of mass and acceleration.

  Force  = mass x acceleration

Now, insert the parameters and find the unknown;

  5000 = 2000 x acceleration

   Acceleration  = \frac{5000}{2000}   = 2.5m/s²

7 0
3 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
The number of mushrooms needed by a pizza restaurant is given by the
kupik [55]

Answer: 24

Explanation:

Given the following equation:

g=3n

Where g is the number of mushrooms in a pizza and n the number of pizzas.

If we know the restaurant will make 8 pizzas (n=8), then:

g=3(8)

g=24 This is the needed number of mushrooms for 8 pizzas

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3 years ago
Tire or false
spayn [35]
1 and 4 are tire.
2 and 3 are not.
8 0
3 years ago
An electric heater rated 600 w operates 6 hours per day find the cast to operate it for 30 days , at rs. 4.00 per unit​
djverab [1.8K]

Answer:

Rs. 432*10^3 (In kilowatts per hour)

I hope it will be useful.

3 0
3 years ago
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