Answer:
Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(
Explanation:
Speed = (wavelength) x (frequency
Speed = (.020 m) x (5 / sec)
Speed = 0.1 m/s
Answer:
The current is halved
Explanation:
The relationship between the current and the resistance is given by Ohm's Law, as follows:

where,
V = Voltage
I = Current
R = Resistance
Therefore, if we double the resistance:

Hence the correct option is:
<u>The current is halved</u>
Answer:
the needle will direct its North South according to the magnetic field of current carrying wire.
Explanation:
A current carrying wire always has a magnetic field around it, in circular loops. This magnetic field will be either clockwise or anticlockwise depending on the direction of current.
Right hand rule tells the direction. Place the current carrying wire in your right hand with thumb pointing the direction of current. Curl of the fingers tell the direction of current.
When the needle gets in the vicinity of the field, its poles aligns itself with the field. (previous position of the compass needle has no effect on its position in the field). The north pole and south pole will be set in the direction of magnetic field.
The distance between the needle and wire does effect the strength (accuracy) of the needle position. Strong field will create strong deflection of the needle whereas when the distance from wire increases, field weakens, thus the deflection of needle will be weak.
Answer:

Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
and
as the initial speed and angle, then we have




If we want to know the maximum height reached by the object, we find the value of t when
becomes zero, because the object stops going up and starts going down

Solving for t

Then we replace that value into y, to find the maximum height

Operating and simplifying

We have

The maximum height is

