Answer:
m = B²qR² / 2 V
Explanation:
If v be the velocity after acceleration under potential difference of V
kinetic energy = loss of electric potential energy
1/2 m v² = Vq ,
v² = 2 Vq / m ----------------------- ( 1 )
In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force
magnetic force = centripetal force
Bqv = mv² / R
v = BqR / m
v² = B²q²R² / m² ------------------------- (2)
from (1) and (2)
B²q²R² / m² = 2 Vq / m
m = B²q²R² / 2 Vq
m = B²qR² / 2 V
It's not true.
The crucial principle for a scientific experiment is to keep only ONE variable at a time.
In this case, the variable of this experiment is actually the tomato is in sunny part or in shady part, instead of whether applying Ca fertilizer.
Answer:
m = 5.22 kg
Explanation:
The force acting on the bucket is 52.2 N.
We need to find the mass of the bucket.
The force acting on the bucket is given by :
F = mg
g is acceleration due to gravity
m is mass

So, the mass of the bucket is 5.22 kg.
Explanation:
Given that,
The initial velocity of a skater is, u = 5 m/s
She slows to a velocity of 2 m/s over a distance of 20 m.
We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

a = acceleration

So, her acceleration is
and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.
In some unusual applications of unusual components, I can think of unusual electric circuits where a switch may be connected in parallel with a device in order to control it.
But I'm sure this is not what's intended in a question on the high-school level.
Until you get in a situation with tricky applications in a tricky circuit, your switches will always be connect <em>in series</em> with the devices they control.