Answer:
Rolling friction is much smaller than sliding friction because Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.
Rolling Friction example: Anything with weels (cars,skateboards) or a ball rooling.
Sliding Friction example: Bicycle brakes,skinning your knee walking,writing.
Answer:
True
Explanation:
Modern safer and cheaper nuclear reactors can not only meet the range of our long term energy demands, they can also fight global warming.
Modern techniques provide ways to reduce radioactive waste amount. "A closed fuel cycle may be switched on for new kinds of nuclear plants. Alternatively, the waste is chemically dissuaded to transform the reusable element into fuel. This implies that nuclear waste would not be buried.
Answer:
Explanation:
F = ma. For us, this looks like
60 = 30a and
a = 2 m/s/s
If the force goes up to, say, 90, then
90 = 30a and
a = 3...if the force goes up, the acceleration also goes up.
If the mass goes up to say, 60, and the force stays the same, then
60 = 60a and
a = 1...if the mass goes up, the acceleration goes down.
Answer:4. Two charged objects have a repulsive force of 0.080 N. If the distance separating the objects is tripled, then what is the new force? Explanation: The electrostatic force is inversely related to the square of the separation distance.
Explanation:
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
