The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
In order to answer this question ... strange as it may seem ...
we only need one of those measurements that you gave us
that describe the door.
The door is hanging on frictionless hinges, and there's a torque
being applied to it that's trying to close it. All we need to do is apply
an equal torque in the opposite direction, and the door doesn't move.
Obviously, in order for our force to have the most effect, we want
to hold the door at the outer edge, farthest from the hinges. That
distance from the hinges is the width of the door ... 0.89 m.
We need to come up with 4.9 N-m of torque,
applied against the mechanical door-closer.
Torque is (force) x (distance from the hinge).
4.9 N-m = (force) x (0.89 m)
Divide each side by 0.89m: Force = (4.9 N-m) / (0.89 m)
= 5.506 N .
Answer:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:
x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s
hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.
Explanation:
Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.
The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.
Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill
