Answer:
12.7m/s
Explanation:
Given parameters:
Mass of the diver = 77kg
Height = 8.18m
Unknown:
Speed of the diver before hitting the water = ?
Solution:
The speed of the diver before hitting the water is the final velocity.
To solve this problem, we use the expression below;
v² = u² + 2gH
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
H is the height
Now insert the parameters and solve;
v² = 0² + 2 x 9.8 x 8.18
v² = 160.328
v = 12.7m/s
Explanation:
It is given that,
Mass of the ball, m = 0.06 kg
Initial speed of the ball, u = 56 m/s
Final speed of the ball, v = -34.5 m/s (opposite direction)
(a) Let J is the impulse delivered to the ball by the racquet. It is equal to the change in momentum of the object as :
J = -5.43 kg-m/s
(b) The work done by the racquet on thee ball is equal to the change in kinetic energy as :
W = -58.372 Joules
Answer:
c. that light can travel in a vacuum
Heat flux (i) = k.a.temp variation/d
i = 250 J/s or 250w ; d = 2.1 x 10^-3m ; A = 1.9m^2 ; K (Body Fat) = 0.2W/m.Kelvin ; Temp. Var. (Delta T) = ?
-> 250 = 0.2 x 1.9 x deltaT/2.1 x 10^-3
-> 250 x 2.1 x 10^-3 = 0.38 x deltaT
-> 525 x 10^-3 = 38x10^-2 x deltaT
-> deltaT = 525 x 10^-3/ 38 x 10^-2
-> deltaT = 13.81 x 10^(-3 - (-2))
-> deltaT = 13.81 x 10^-1
-> deltaT = 1.381 K
Answer:
The two general types of air pollution are _____ and _____.
answer: gases and particles
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