Answer:
Tires.
Explanation:
There are the few steps which are discussed below should be taken to increase or extend the life of tires.
(1) Avoid fast starts: Fast start of the vehicle will increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.
(2) Avoid fast stop: Fast stop of the vehicle will also increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.
(3) Avoid sharp turns: The alignment of the wheels and tires are in such a way that they work properly when vehicle is drive in a straight path but sharp turn will increase the uneven pressure on the tires will lead to decrease the life of tires.
Therefore, the life of tires can be extend by avoiding all the above mention actions such as fast stop, start and sharp turns.
Answer:
1.4 m
Explanation:
v = Speed of sound in water = 1400 m/s
f = Frequency of sound = 1000 Hz
= Wavelength
When we multiply the frequency and the wavelength of a wave we get the velocity of sound in that medium

The wavelength of the sound waves in water is 1.4 m
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is
.
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '
' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is

Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

Part(c):
If we apply Gauss' law of electrostatics, then

The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
