Answer:
80Ω
Explanation:
Resistance in a series circuit (when one resistor is after another in the same line) is calculated with the following equation
Total resistance = R1 + R2 + R3...Rn
Substitute in your values and solve for total resistance
Total resistance = 60 + 20 = 80Ω
the team rotates the serve
Answer:
Electric field acting on the electron is 127500 N/C.
Explanation:
It is given that,
Mass of an electron, 
Charge on electron, 
Initial speed of electron, u = 0
Final speed of electron, 
Distance covered, s = 2 cm = 0.02 m
We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :
According to Newton's law, force acting on the electron is given by :
F = ma
Electric force is given by :
F = q E, E = electric field
E = 127500 N/C
So, the electric field is 127500 N/C. Hence, this is the required solution.
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:
If a crest formed by one wave interferes with a trough formed by the other wave then the rope will not move at all.
Explanation:
Assume a straight rope tied to both ends is at rest. When a wave is created at one end of the rope, it travels to the other end of the rope through formation of alternative crest and trough. Due to these crest and trough the rope shifts up and down.
But when there are two waves travelling through the rope and both have opposite direction (directed towards one another) in such a way that crest formed by one wave is interfering with the trough formed by the other wave then due to this interference the waves will cancel the effects of each other on the rope and rope will be stable.
