I think its c or d but im not quite sure.....
You'll hear that force called different things in different places. It
may be called "electromotive force", "EMF", "potential difference",
or "voltage".
It's just a matter of somehow causing the two ends of the wire
to have different electrical potential. When that happens, the
free electrons in the copper suddenly have a burning desire to
travel ... away from the end that's more negative, toward the end
that's more positive, and THAT's an "electric current".
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;

The steepness of the slope is therefore;

Where;
= Half the width of the truck =
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m



The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
Learn more here:
brainly.com/question/20793607
Answer:
The magnitude of the rate of change of the child's momentum is 794.11 N.
Explanation:
Given that,
Mass of child = 27 kg
Speed of child in horizontal = 10 m/s
Length = 3.40 m
There is a rate of change of the perpendicular component of momentum.
Centripetal force acts always towards the center.
We need to calculate the magnitude of the rate of change of the child's momentum
Using formula of momentum


Put the value into the formula


Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.
Explanation:
They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.
(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)
In this case, you can't ignore friction because the disk wouldn't roll without it.
As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.
(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)
Hopefully this helps clear up the misunderstanding for you.