Answer: D) Earthquake
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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:
The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg
Explanation:
From the formulae
F = qvB and F = mv²/r
Where F is Force
q is charge
v is speed
B is magnetic field strength
m is mass
and r is radius
Then,
qvB = mv²/r
qB = mv/r
We can write that
q/m = v/rB ---- (1)
Also
From Electric force formula
F = Eq
Where E is the electric field
and magnetic force formula
F = Bqv
Since, electric force = magnetic force
Then, Eq = Bqv
E = Bv
∴ v = E/B
Substitute v = E/B into equation (1)
q/m = (E/B)/rB
∴ q/m = E/rB²
(NOTE: q/m is the charge to mass ratio)
From the question,
E = 3.10 ×10³ N/C
r = 4.20 cm = 0.0420 m
B = 0.360 T
Hence,
q/m = 3.10 ×10³ / 0.0420 × (0.360)²
q/m = 569517.9306 C/kg
q/m = 5.7 × 10⁵ C/kg
Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.
Answer:
Explanation:
Given
mass of each car
acceleration of train
Force required to pull this system
For first car
for second car
this form a pattern of the form
for (a)30 th and 31 st car tension
For 49 th and 50 th car