Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
<em />
They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
3.13 m/s2
.
the formula for acceleration is as follows:
force/mass = acceleration
-
so 25/8 = 3.13
1. 
Explanation:
We have:
voltage in the primary coil
voltage in the secondary coil
The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.
2. 5.7 V
We can solve the problem by using the transformer equation:

where:
Np = 400 is the number of turns in the primary coil
Ns = 19 is the number of turns in the secondary coil
Vp = 120 V is the voltage in the primary coil
Vs = ? is the voltage in the secondary coil
Re-arranging the formula and substituting the numbers, we find:

<h2>~<u>Solution</u> :-</h2>
- Here, the <u>moment arm</u> is defined as follows;
The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.
