What is the difference between the fields surrounding a stationary charge and those that surround a moving charge? A moving charge has only an electric field and a stationary charge has only a magnetic field. A moving charge has both an electric and magnetic field and a stationary charge has only an electric field. A moving charge has only a magnetic field and a stationary charge has both an electric and a magnetic field. There is no difference between the fields surrounding a stationary charge and those that surround a moving charge—both have an electric and a magnetic field.

Answers

Answer 1

Answer: The best statement would be the second option where a moving charge has both an electric and magnetic field and a stationary charge has only an electric field. A charged particle that is stationary produces a magnetic field but does not produce electric field. A periodically moving charge, on the other hand produces an electro-magnetic field.

Answer 2

Answer:

The correct answer to the question is : B) A moving charge has both an electric and magnetic field, and a stationary charge has only an electric field.

EXPLANATION :

Before answering this question, first we have to understand the nature of field created by charge particles.

If the charge particle is at rest i.e stationary, then the field created by the charge particle is electric in nature. Hence, a stationary charge particle produces electric field only.

If a charge particle moves with uniform velocity, then the field will be only magnetic .Hence, a non accelerating charge particle will produce magnetic field.

If a charge particle accelerates, then the field created by the charge particle is electromagnetic in nature.

Hence, the correct answer of this question is the second statement.

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88.0 km to m (in physics prefixes )

Answers

We know that,

kilometre = 1000 metre

1000 can be written as 10³ which is kilo.

Given,

88.0km = 88.0×1000

= 88.0×10³m

An 89 kg person climbs up a uniform 13 kg ladder. The ladder is 5.9 m long; its lower end rests on a rough horizontal floor (static friction coefficient 0.26) while its upper end rests against a frictionless vertical wall. The angle between the ladder and the horizontal is 52◦. Let d denote the climbing person's distance from the bottom of the ladder (see the above diagram). When the person climbs too far (d > dmax), the ladder slips and falls down (kaboom!). Calculate the maximal distance dmax the person will reach before the ladder slips. The acceleration of gravity is 9.8 m/s2.

Answers

Answer:

The maximum distance the person will reach before he slip is 1.82m.

Not even close to the middle of the ladder.

Explanation:

Check attachment for solution

At which point on this electric field will a test charge show the maximum strength?A
B
C
D

Answers

Answer:

Explanation:

The figure shows the electric field produced by a spherical charge distribution - this is a radial field, whose strength decreases as the inverse of the square of the distance from the centre of the charge:

$E\propto (1)/(r^2)$

More precisely, the strength of the field at a distance r from the centre of the sphere is

$E=k(Q)/(r^2)$

where k is the Coulomb's constant and Q is the charge on the sphere.

From the equation, we see that the field strength decreases as we move away from the sphere: therefore, the strength is maximum for the point closest to the sphere, which is point A.

This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.

Jason hits a volleyball with an initial velocity of 6 meters per second straight up. if the back starts at 2 meters above the floor how long will the ball be in the air

Answers

Vi = 6 m/s
d = -2 m
a = - 9.81 m/s^2
t = ?
Formula
d = vi*t + 1/2 a t^2
- 2 = 6m/s *t - .5 * 9.81 * t^2
Put the - 2 on the right.
4.905 t^2 + 6 m/s t +2 = 0
a = -4.905 m/s^2
b = 6 m/s
c = 2 m
t = [- b +/- square root(b^2 - 4ac)]/(2a)
t = [ -6 +/- square root (6^2 - 4*(-4.905)*2)] / (- 4.905*2)
t = [-6 +/- 8.674) / - 9.81
t = -14.674/-9.81
t = 1.69 seconds.

Can someone help with number 4 please

Answers

The way I look at these things is like this:

-- The runner covered 12 meters in 4 seconds.
Average speed = 3 meters per second.

-- Speed at the beginning = zero.
In order to make the average 3, Speed at the end = 6 meters per second.

-- Speed increased from zero to 6 meters per second in 4 seconds.
It must have increased 1.5 meters per second each second.

That's choice-#2.

Here we have -
Acceleration (a) = ?
Initial velocity (u) = 0 m/s (runner starts from rest)
Distance (s) = 12 m
Time (t) = 4 seconds

So, using the second equation of motion, we have-

$s = ut + (1)/(2)at^(2)$

$12 = 0*4 + (1)/(2)*a*4 ^(2)$

$12 = 0+8a$

$(12)/(8) = a$

$1.5m/ s^(2)$

So, option (2) is correct.

An electrical kettle uses 18A of current to boil 500ml of water. What is the electrical resistance of the kettle in ohms?

Answers

We can't tell from the given information.

Regarding the water, we don't know . . .
-- starting temperature
-- quantity of water
-- time in which the heater boiled it
so we don't know how much heat energy it takes to boil it.

Even if we DID know the total heat energy required, we don't know
the voltage supplied to the kettle's heating coil. Together with the
current flowing in the heater, we need to know either its resistance
or the voltage across it in order to calculate the rate at which heat
energy is dissipated from the heater.

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