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.
Answer:
The observed wavelength on Earth from that hydrogen atom is
.
Explanation:
Given that,
The actual wavelength of the hydrogen atom, 
A hydrogen atom in a galaxy moving with a speed of, 
We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

is the observed wavelength

So, the observed wavelength on Earth from that hydrogen atom is
. Hence, this is the required solution.
Answer: K.E = 0.4 J
Explanation:
Given that:
M = 1.0 kg
h = 0.04 m
K.E = ?
According to conservative of energy
K.E = P.E
K.E = mgh
K.E = 1 × 9.81 × 0.04
K.E = 0.3924 Joule
The kinetic energy of the pendulum at the lowest point is 0.39 Joule
An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”
Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”
Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”
Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.
Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)
Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?
You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.
Hope this helps!
Answer:

Explanation:
The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the
resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:
