The answer is 0k because 143c equals nothing
Earths Rotation
Earth's rotation is the rotation of Planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north<span> pole star Polaris, Earth turns counter</span>clockwise<span>. </span>
"AB84"
It's either 3 or 4 I know this becuase I have read a book about electricity
<span>Answer:
The moments of inertia are listed on p. 223, and a uniform cylinder through its center is:
I = 1/2mr2
so
I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2
Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration:
t = Ia
-1.20 Nm = (0.0120984 kgm2)a
a = -99.19 rad/s/s
Now we have a kinematics question to solve:
wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s
w = 0
a = -99.19 rad/s/s
Let's find the time first:
w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s
t = 10.558 s = 10.6 s
And the displacement (Angular)
Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form
s = (u+v)t/2
Which is
q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s
q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians
But the problem wanted revolutions, so let's change the units:
q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
Answer:
9.43 m/s
Explanation:
First of all, we calculate the final kinetic energy of the car.
According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:
where
W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)
is the final kinetic energy
is the initial kinetic energy
Solving,
Now we can find the final speed of the car by using the formula for kinetic energy
where
m = 661 kg is the mass of the car
v is its final speed
Solving for v, we find
