Answer
given,
position of particle
x(t)= A t + B t²
A = -3.5 m/s
B = 3.9 m/s²
t = 3 s
a) x(t)= -3.5 t + 3.9 t²
velocity of the particle is equal to the differentiation of position w.r.t. time.
------(1)
velocity of the particle at t = 3 s
v = -3.5 + 7.8 x 3
v = 19.9 m/s
b) velocity of the particle at origin
time at which particle is at origin
x(t)= -3.5 t + 3.9 t²
0 = t (-3.5 + 3.9 t )
t = 0, 
t = 0 , 0.897 s
speed of the particle at t = 0.897 s
from equation (1)
v = -3.9 + 7.8 t
v = -3.9 + 7.8 x 0.897
v = 3.1 m/s
The force needed to the stop the car is -3.79 N.
Explanation:
The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.
So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be
Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,
Work done = Final kinetic energy - Initial kinetic energy
Work done = Force × Displacement
So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.
Thus, the force needed to the stop the car is -3.79 N.
M=meter, km=kilometer, mm=millimeter, mg=micrometer, cm=centimeter
Answer:
Explanation:
Moment of inertia of each wheel = 1/2 m R²
m is mass and R is radius of wheel
= .5 x 15.5 x .175²
= .2373 kg m²
moment of inertia of tyre
1/2 m ( r₁² + r₂² )
= 1/2 x 1.9 x ( .315² + .19²)
= 1/2 x 1.9 x ( .099+ .036)
= .12825 kg m²
moment of inertia of tread
= 1/2 m r²
= .5 x 12 x .335²
= .67335 kg m²
moment of inertia of axle
= 1/2 m r ²
= .5 x 14.5 x .0195²
= .00275
moment of inertia of drive shaft
= 1/2 x 32.5 x .029²
= .0137 kg m ²
Total moment of inertia of one tyre
= 1.05535 kg m²
total moment of inertia of two rear wheels
= 2.1107 kg m²
95 % of torque
= .95 x 180
= 171 Nm
angular acceleration
= torque / moment of inertia
= 171 / 2.1107
= 81.01 radian /s²
Answer:
2,8,8
Explanation:
im not sure really what you asking but is it this the shells
