Answer:
when a force is applied by one object to a second object, an equal and opposite force is applied back on the first object
Explanation:
Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
Answer:
True
Explanation:
Because in atom the negative charge become lose or gain.
Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = 
Using equation to solve the problem

E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water



now ,


taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)