Gases do not conduct heat well.
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
Answer:
option (E) 1,000,000 J
Explanation:
Given:
Mass of the suspension cable, m = 1,000 kg
Distance, h = 100 m
Now,
from the work energy theorem
Work done by the gravity = Work done by brake
or
mgh = Work done by brake
where, g is the acceleration due to the gravity = 10 m/s²
or
Work done by brake = 1000 × 10 × 100
or
Work done by brake = 1,000,000 J
this work done is the release of heat in the brakes
Hence, the correct answer is option (E) 1,000,000 J
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
