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
Answer: 20.4752789138x x 10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x 10^23 atoms= 20.4752789138x x 10^23 atoms
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
Learn more about tangential velocity here: brainly.com/question/25780931
Answer:
Gravity: downwards
Air drag and air-pressure on the inner surface of the the parachute: Upwards
Explanation:
- If a sky-diver is in the final stages of his descend with open parachute such that the wind is calm and it does not blows him laterally.
- In such a condition the air resistance in the form of drag and the pressure force due to the air captured in the parachute are acting in the upward direction which balance the force of gravity on the body. But this situation may occur momentarily and then again the diver must begin to slowly descend.
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
