... 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 : 
Explanation :
It is given that,
Mass of the engine, m = 30 kg
Thrust is equivalent to the force acting perpendicularly and it is F = 300 N
According to Newton's second law of motion :

a is the acceleration of the engine.



So, the acceleration of the engine is
.
Hence, this is the required solution.
To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

Where
Angular acceleration
Angular velocity
t = Time
Our values are




Replacing at the previous equation we have that the angular velocity is



Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be



Therefore the angular acceleration of a point on the outer edge of the tires is 
A. The particles are packed more tightly in materials with more density which causes the vibrations to bounce of the partials more rapidly which makes them go faster
Answer:
0.707m
Explanation:
from formula of range i.e R=Usin2Q/g