Answer:
I already told ya
Explanation:
i got brainly so i could learn
Answer:
(A⃗ ×B⃗ )⋅C⃗ = 69.868
Explanation:
We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment
When a body strictly moves on a curve, it's velocity at a point is tangential to the curve at that point.
Centripetal acceleration is the acceleration that a body experiences by the virtue of change in it's tangential velocity. It is directed towards the centre and mathematically is v^2/R where v is the speed at the instant.
So, 18 = v^2/R
v^2 = 504
v = 6√14
Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.
</em>
<em>
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
The amount of work done by the motor is given by:


Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

=> 
Now we can calculate the final velocity:

Finally, we calculate the total work:

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J
Answer:
As 28m/s = 28m/s
Explanation:
r = the radius of the curve
m = the mass of the car
μ = the coefficient of kinetic friction
N = normal reaction
When rounding the curve, the centripetal acceleration is

therefore



As 28m/s = 28m/s