What is DC in electricity mean?

Dahasolnce [82]

Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work.

So they are both closely related to each other.

HOPE THIS HELPS

3 0

3 years ago

Why is it warmer near the equator than near the poles?

maw [93]

The equator is warmer than the poles because the equator is closer to the sun. In other words, the sun is overhead the equator, which is a result of the Earth's curvature.

3 0

3 years ago

A farmhand attaches a 25-kg bale of hay to one end of a rope passing over a

Georgia [21]

Explanation:

Newton's second law:

∑F = ma

277 N − 245 N = (25 kg) a

a = 1.28 m/s²

8 0

3 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

3 years ago

A ball is released from rest at a height of 10 m and falls freely to the

ElenaW [278]

Answer:

The new kinetic energy would be 16 times greater than before.

Explanation:

Kinetic energy is found using this formula:

KE = 1/2mv²
where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)

We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

√(2gh)

Let's substitute h = 10 m and h = 40 m into this formula.

√(2g(10))
√(2g(40))

We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

6 0

3 years ago