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Agata [3.3K]
3 years ago
10

For a Van de Graaff generator to acquire a charge on its surface what must happen?

Physics
1 answer:
horsena [70]3 years ago
3 0
The answer is D. A Van de Graaff generator is an electrostatic generator which utilizes a moving belt to aggregate an electric charge on an empty metal globe on the highest point of a protected section, making high electric possibilities. It delivers high voltage coordinate current (DC) power at low current levels.
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Sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motionstays in motion with the sam
Lera25 [3.4K]

Answer:

D

Explanation:

Newtons first law states that if an object is at rest it will stay at rest only if an unbalanced force acts on it. As well as if an object is in motion it will stay in motion unless an unbalanced force acts on it.

Ps- The object will stay moving in the same speed and direction.

5 0
3 years ago
Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth
barxatty [35]
An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F).  An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0
4 years ago
A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr
kolbaska11 [484]

Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

4 0
3 years ago
A high diver dives into a swimming pool. His potential energy at the top is 10,000 J (relative to the surface of the pool). What
Semmy [17]

Answer:

Kinetic energy of diver at 90% of the distance to the water is 9000 J

Explanation:

Let d is the distance between the position of the diver and surface of the pool.

Initially, the diver is at rest and only have potential energy which is equal to 10000 J.

As the diver dives towards the pool, its potential energy is converting into kinetic energy due to law of conservation of energy, as total energy of the system remains same.

Energy before diving = Energy during diving

(Potential Energy + Kinetic Energy) =  (Kinetic Energy +  Potential Energy)

When the diver reaches 90% of the distance to the water, its kinetic energy

is 90% to its initial potential energy, as its initial kinetic is zero,i.e.,

K.E. = \frac{90}{100}\times10000

K.E. = 9000 J

6 0
3 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
4 years ago
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