Answer:
26.6 m/s
Explanation:
Given:
Δy = 2.1 m
t = 5.35 s
a = -9.8 m/s²
Find: v₀
Δy = v₀ t + ½ at²
(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²
v₀ = 26.6 m/s
Answer:
The position of the image from the camera lens is approximately 121.1 mm
Explanation:
The given parameters of the lens are;
The specification of the camera lens = 120 mm
Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m
The distance of the object from the camera, = 13 m
The lens equation for finding the position of the image is given as follows;
Where;
= The position of the image from the camera lens
Therefore, by plugging in the known values, we have;
The position of the image from the camera lens, = 0.1211 m = 121.1 mm
Answer:
exothermic
Explanation:
energy is absorbed by the surroundings
Answer:
the initial velocity of the ball is 104.67 m/s.
Explanation:
Given;
angle of projection, θ = 60⁰
time of flight, T = 18.5 s
let the initial velocity of the ball, = u
The time of flight is given as;
Therefore, the initial velocity of the ball is 104.67 m/s.