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Nitella [24]
2 years ago
13

In the presence of a dielectric, the capacitance of a electric field inside the plates now is: a) Less b) More c) Same as the el

ectric field in absence of the dielectric d) Zero
Physics
1 answer:
Nesterboy [21]2 years ago
6 0

Answer:

Explanation:

As the dielectric is inserted between the plates of a capacitor, the capacitance becomes K times and the electric field between the plates becomes 1 / K times the original value. Where, K be the dielectric constant.

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Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from
mojhsa [17]

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

8 0
3 years ago
A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors
ale4655 [162]

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, F_1 = 140 N

let the second force = F_2

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N

F_2 = 280 \ East

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N

F_2 = 560 \ West

8 0
3 years ago
2 forces are acting on a body. I forgot the body doesn't it's position or shape, the forces.......a)must be of equal magnitude b
Veseljchak [2.6K]

Answer:

C

Explanation:

only if there is a net force of zero, the body will not move

some people may say B but that is wrong because maybe one force is greater than the other so the object would still move even though the forces are in opposite directions and parallel

5 0
3 years ago
Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a
blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force F_c. In this case, the only force that applies for that is the static frictional force f_sbetween the tires and the track. Then, we can write that:

f_s=F_c

And since f_s\leq \mu N and F_c=\frac{mv^{2}}{r}, we have:

\mu N\geq \frac{mv^{2}}{r}

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

N-mg=0\\\\\\implies N=mg

Substituting this expression for N and solving for v, we get:

\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

7 0
2 years ago
Read 2 more answers
1) Calculate the potential energy of a 5.00 kg object sitting on a 3.00 meter high ledge.
maria [59]

Answer:

15kg

Explanation:

5 0
2 years ago
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