All categories

3 years ago

Repulsion is the sure test of electrification.explain.

1 answer:

3 0

Attraction may occur between two oppositely charged bodies or a charged body and an uncharged one. That is why in attraction, one is not sure if both bodies are charged, However in repulsion, there are two bodies that have the same charge. That is why repulsion is the sure test of electrification.

You might be interested in

Which of the following can be done to demagnetize a permanent magnet?

ICE Princess25 [194]

Place the permanent magnet into an AC magnetic field and slowly reduce the field to 0 and the magnet will be temporarily demagnetized.

6 0

4 years ago

A 40 kg shell is flying at a speed of 72 km/h. It explodes into two

viva [34]

Answer: This is true

Explanation: v2=800/25= 32m/s

5 0

3 years ago

If the same pressure is exerted over a greater area will more of less force result? Group of answer choices :

Zigmanuir [339]

Answer:

More force

Explanation:

Pressure and force are related by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area

We can re-arrange the equation as

$F=pA$

In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.

8 0

4 years ago

In a process called pair production, an energetic

vredina [299]

Answer:

(3) mass-energy must be conserved

Explanation:

As we know that gamma rays are mass less and charge less photons which will have sufficient energy.

Now in case of Pair production the gamma photons convert its whole energy into mass by the law of Einstein's mass energy equivalence relation.

As per his relation we can say

$E = \Delta m c^2$

here we will have

[tex]\Delta m[\tex] = mass produced

now we also have to think that here the two particles must have a pair of particles and antiparticles so that the combined mass system will have energy equivalent to the energy of gamma photons and also it must follow the conservation of charge

So here it will form an electron and a positron such that total charge will be zero and the energy will be same as energy of gamma photon.

3 0

4 years ago

Read 2 more answers

You’re driving your car towards an intersection. A Porsche is stopped at the red light. You’re traveling at 36 km/h (10 m/s). As

34kurt

Your position at time $t$ , relative to the stop line:

$x_1=-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t$

The Porsche's position:

$x_2=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2$

a. You pass the Porsche immediately after the time it takes for $x_1=x_2$ :

$-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=2.3\,\rm s$

at which point you both will have traveled 7.8 m from the stop line.

b. The equation in part (a) has two solutions. The Porsche passes you at the second solution of about $t=4.4\,\rm s$ , at which point you both will have traveled 29 m.

c. At time $t$ , the Porsche is moving at velocity

$v=\left(3\dfrac{\rm m}{\mathrm s^2}\right)t$

so that at the moment it passes you, its speed is 13 m/s, which is about 46.8 km/h and below the speed limit, so neither of you will be pulled over.

6 0

4 years ago