Answer:
72 volts.
Explanation:
To solve this, we have to use the Ohm's law.
The ohm's law tells us that the voltage drop of a resistor is directly proportional to the current applied to the conductor.
in this case the current is 1.8 amps and the resistor is 40 ohm
so
.
Answer:
The voltage would be 
Explanation:
From the question we are told that
The wavelength is 
Generally when a electron changes state to a higher state an energy(E) is given off and this is equivalent to eV
So the voltage of the electron can be evaluated as
Where h is the planks constant with a constant value 
c is the speed of light 
e is the value charge in one electron 
Substituting value
So the fourth current drop would be recorded when electron has change state four times and the voltage at this point can be mathematically evaluated as
When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.
Answer:
The arrow will leave the bow with a velocity of 10 m/s.
Explanation:
Hi there!
The potential energy stored in the bow can be calculated using the following equation:
U = 1/2 · k · d²
Where
U = elastic potential energy.
k = spring constant.
d = stretched distance of the bow
Then:
U = 1/2 · 112 N/m · (0.29 m)²
U = 4.7 J
When the bow is released, the potential energy is transformed into kinetic energy. Then, the kinetic energy of the arrow when it leaves the bow will be:
KE = 1/2 · m · v² = 4.7J
Where:
KE = kinetic energy.
m = mass of the arrow.
v = velocity of the arrow:
Then:
4.7 J = 1/2 ·0.094 kg · v²
2 · 4.7 J / 0.094 kg = v²
9.4 kg · m²/s² / 0.094 kg = v²
v = 10 m/s
The arrow will leave the bow with a velocity of 10 m/s.
B) the resistor
Because If you were going to connect the (+) and the (-) and put it on the resistor the circuit would work
