Force is a vector quantity & must<span> be treated accordingly; the </span>electric force between two<span> objects always acts along the line that connects their centers of </span>charge<span> ... </span>Electric forces can<span> be </span>either attractive<span> or</span>repulsive<span>, but </span>gravitational forces<span> are always </span>attractive<span> (because objects </span>can<span> have </span>either<span> a positive or a negative </span>charge<span>, .</span>
Answer:..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Volosity???????
Explanation: THINK THATS IT PROBABLY MISPELLED IT SORRY IN ADVANCE IF WRONG
When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.
But linear momentum still has to be conserved. It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.
Momentum = (mass) x (velocity)
After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
(5.5 x 6.8) = 37.4 kg-m/s to the right.
In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.
M x V = (6 x speed) = 37.4 kg-m/s to the left.
Divide each side by 6 : Speed = 37.4 / 6 =<em> 6.2333... m/s left</em>
(That number is (6 and 7/30) m/s .)
The sl unit of work is the joule (j)
Answer:
The intensity of a light from the source at a distance of 10 meters is 0.0026 foot-candle.
Explanation:
Given that,
Distance from the source = 2 meter
Intensity = 0.065 foot -candle
New distance = 10 m
We know that,
The intensity I of light varies inversely with the square of the distance from the source.
We need to calculate the value of constant
Using formula of intensity
Put the value into the formula
We need to calculate the intensity
Using formula of intensity again
Put the value into the formula
Hence, The intensity of a light from the source at a distance of 10 meters is 0.0026 foot-candle.