Answer:
The time is 133.5 sec.
Explanation:
Given that,
One side of cube = 10 cm
Intensity of electric field = 11 kV/m
Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.
We need to calculate the rate of energy transfer from the beam to the cube
Using formula of rate of energy


Put the value into the formula


We need to calculate the amount of heat
Using formula of heat


Put the value into the formula


We need to calculate the time
Using formula of time

Put the value into the formula


Hence, The time is 133.5 sec.
Explanation:
63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:

Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:

Now, we can solve for the period:

The three ways a person can manipulate light
would be the following:,
filter, and the time the photograph is taken
<span>1.
</span>Angle
- <span>The </span>camera angle<span> <span>marks
the specific location at which the movie </span></span>camera<span> <span>or
video </span></span>camera<span> is
placed to take a shot.</span>
<span>2.
</span>Filter - Camera<span> <span>lens </span></span>filters<span> <span>still have many uses in digital photography,
and should be an important part of any photographer's </span></span>camera<span> bag.</span>
<span>3.
</span>Time
the photograph is taken - The golden hour, sometimes called the "magic
hour", is roughly the first hour of light after sunrise, and the last hour
of light before sunset, although the exact duration varies between seasons.
During these times the sun is low in the sky, producing a soft, diffused light
which is much more flattering than the harsh midday sun that so many of us are
used to shooting in.
I am hoping that these answers
have satisfied your queries and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.
The electron is accelerated through a potential difference of

, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:

where
m is the electron mass
v is the final speed of the electron
e is the electron charge

is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:

where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B: