Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x 
= τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N
... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.
... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹ = 5 ohms.
... The 5-ohm-equivalent block and the 20-ohm resistor form a
voltage divider across the battery.
The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .
... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .
... With 0.4A flowing through it, the 9-ohm resistor is dissipating
I²R = (0.4A)² (9 ohms) = (0.16 A²) (9 ohms) = 1.44 W (choice-3)
Answer:
The orbital speed can be found using v = SQRT(G*M/R). The R value (radius of orbit) is the earth's radius plus the height above the earth - in this case, 6.59 x 106 m.
