Answer:
the sheets approach while the object is near
Explanation:
An electroscope is an apparatus that has two metal sheets attached, when these sheets have a charge and distribute evenly between them and the sheets repel.
When I approach an object charged with a counter (negative) charge, part of the charge of the electroscope moves near the charged external object, to neutralize the electric field, so as the charge on one of the plates decreases the electroscope has approached , as the objects are not touched the system remains in this configuration while the object is close. When the object is released, the electric field it creates disappears, so the positive charges repel inside the electroscope and the sheets repel to the initial position.
In short, the sheets approach while the object is near
If the bulb is in series with something else, then . . .
-- The brightness of the bulb depends on the <em>other</em> device in the circuit.
-- If the other device is designed to use <em>less power</em> than the bulb, then the
other device gets <em>more power</em> than the bulb gets.
-- If the other device is designed to use <em>more power </em>than the bulb, then the
other device gets <em>less power</em> than the bulb gets.
-- If the other device is removed from the circuit, then the bulb doesn't light at all.
This description of the often-screwy behavior of a series circuit may partly explain
why the electric service in your home is not a series circuit.
Answer:
The time interval is
Explanation:
From the question we are told that
The constant acceleration is
The displacement is
According to the second equation of motion we have that
given that the blade started from rest
which is the initial angular velocity is 0
So
=>
substituting values
=>
=>
Answer:
It is called force of friction
Explanation:
The force of friction is a force that acts between two objects whose surfaces are in contact with each other.
Consider the typical case of an object sliding along a certain surface. There are two types of frictions:
- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:
However, this force of static friction has a maximum value, which is given by
where
is the coefficient of static friction
N is the normal reaction exerted by the surface on the object
So, when becomes greater than , the static friction is no longer able to balance the push applied, and the object will start sliding forward.
- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by
where
is the coefficient of kinetic friction, and its value is generally smaller than . The direction of this force is also opposite to the direction of motion of the object.
a) 5.0 m/s
This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:
where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.
When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:
where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find
b) 4.0 m/s
After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:
where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find
c) 2.8 m
We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:
While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:
where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find