You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.
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
Either theory or evidence
Answer: A) 2 B) 4 C) 1
Explanation:
The Electric field from a parallel-plate capacitor is given by:
A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.
B) from the above expression for the electric field, If the size of the plate is double, then the E final is four times weaker that the original.
C) If the distante between plates is doubled the final electric field is the same that initial.
When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat
So here weight of the boat = Buoyancy force
let say boat is sunk by distance "h"
now we can say
now by above force balance equation we can write
so boat will sunk by total 5 mm distance
