Find the acceleration of a body whose velocity increases from 11ms-1 to 33ms-1 in 10 seconds

The well-known equation... E = m c²... does not address the origin of the mass involved.

Converting 1 kg of any mass entirely into energy generates

(1kg) · (c²) Joules of energy.

E = (1 kg) · (c²) = (1 kg) · (299,792,458 m/s)²

E = 8.9876 x 10¹⁶ Joules

To simplify, this equates to the energy needed to keep a 100-watt light bulb illuminated for about 10,402,259,010 days.

(This is roughly 28.5 million years, based on the current understanding of days and years.)

Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

Input power = 6 Watts

To find:

What is the output power?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss corresponds to a ratio of 0.5

Output power can be calculated as follows:

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watts

Therefore, the output power of the circuit is 3 Watts.

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

where

To find the average velocity, we divide the displacement by the elapsed time:

At time t = 0, the position is:

At time t = 2.00 s, the position is:

This leads us to the displacement of

The duration for this interval is

Therefore, the average velocity during this period is

b)

At time t = 0, the position is:

At time t = 4.00 s, the position is:

Thus, the displacement is

The time interval is

This yields an average velocity of

c)

The position at t = 2 s is:

And at t = 4 s it is:

This gives us a displacement of

While the time interval is

So the resulting average velocity is

Find out more about average velocity:

B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.

Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.