This affirmative is false
Answer:
0.00970 s
Explanation:
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
Force due to magnetic field = qvB sin θ
q = charge on the particle = 5.4 μC
v = velocity of the charge
B = magnetic field strength = 2.7 T
θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1
F = qvB
Centripetal force responsible for circular motion = mv²/r = mvw
where w = angular velocity.
The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field
mvw = qvB
mw = qB
w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)
w = 3.24 × 10² rad/s
w = 324 rad/s
w = (angular displacement)/time
Time = (angular displacement)/w
Angular displacement = π rads (half of a circle; 2π/2)
Time = (π/324) = 0.00970 s
Hope this Helps!!!
Answer:
Answer for A
Explanation:
F1=GmM/r1^2
If r2 becomes r2=5r
F2=GmM/(25r^2)
Multiply with 25 gives to maintain the same force
I.e.,25F2=F1
F2=G(25m)M/25r^2=F1
By the factor 25 would change to increase to same.
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec
Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
