Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s
Answer:
Δx = 6.33 x 10⁻³ m = 6.33 mm
Explanation:
We can use the Young's Double Slit Experiment Formula here:

where,
Δx = distance between consecutive dark fringes = width of central bright fringe = ?
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
L = distance between screen and slit = 3.7 m
d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m
Therefore,

<u>Δx = 6.33 x 10⁻³ m = 6.33 mm</u>

- Initial velocity,u = -2 m/s
- Final velocity,v = -10 m/s
- Time taken, t = 4 seconds

Find the acceleration ( a ) .

We know that,

Substituting the values in the above formula, we get




Hence,the acceleration of a body is -2 m/s².
I think the main component is iron