The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.
Answer: Option B
<u>Explanation:
</u>
Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.
Answer:




Explanation:
r = Radius of disk = 7.9 cm
N = Number of revolution per minute = 1190 rev/minute
Angular speed is given by

The angular speed is 
r = 2.98 cm
Tangential speed is given by

Tangential speed at the required point is 
Radial acceleration is given by

The radial acceleration is
.
t = Time = 2.06 s
Distance traveled is given by

The total distance a point on the rim moves in the required time is
.
<span>1078 kgm / s would be the answer I hope this helps!!!</span>
Your answer is "<span>surface of a sphere"
Hope this helps.</span>
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>