Impulse = (force) x (length of time the force lasts)
I see where you doodled (60)(40) over on the side, and you'll be delighted
to know that you're on the right track !
Here's the mind-blower, which I'll bet you never thought of:
On a force-time graph, impulse (also change in momentum)
is just the <em>area that's added under the graph during some time</em> !
From zero to 60, the impulse is just the area of that right triangle
under the graph. The base of the triangle is 60 seconds. The
height of the triangle is 40N . The area of the triangle is not
the whole (base x height), but only <em><u>1/2 </u></em>(base x height).
1/2 (base x height) = 1/2 (60s x 40N) = <u>1,200 newton-seconds</u>
<u>That's</u> the impulse during the first 60 seconds. It's also the change in
the car's momentum during the first 60 seconds.
Momentum = (mass) x (speed)
If the car wasn't moving at all when the graph began, then its momentum is 1,200 newton-sec after 60 seconds. Through the convenience of the SI system of units, 1,200 newton-sec is exactly the same thing as 1,200 kg-m/s . The car's mass is 3 kg, so after 60 sec, you can write
Momentum = M x V = (3 kg) x (speed) = 1,200 kg-m/s
and the car's speed falls right out of that.
From 60to 120 sec, the change in momentum is the added area of that
extra right triangle on top ... it's 60sec wide and only 20N high. Calculate
its area, that's the additional impulse in the 2nd minute, which is also the
increase in momentum, and that'll give you the change in speed.
I believe the process you have described is known as an 'experiment'.
Resistance = (voltage) / (current)
For this piece of wire . . .
Resistance = (61 volts) / (6 Amperes)
Resistance = (61/6) (V/A)
<em>Resistance = (10 and 1/6) ohms</em>
Since you know the voltage and current, the length doesn't matter.
Answer:
The magnitude of the radial acceleration is 0.754 rad/s²
Explanation:
Given;
radius of the flywheel, r = 0.2 m
initial angular velocity of the flywheel, 
angular acceleration of the flywheel, a = 0.900 rad/s².
angular distance, θ = 120⁰
the angular distance in radian = 
Apply the following kinematic equation to determine the final angular velocity;
The magnitude of the radial acceleration is calculated as;
Therefore, the magnitude of the radial acceleration is 0.754 rad/s²
