Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.
Answer:
Explanation:
There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman
Taking torque of all forces about the end far away from the woman
Torque by reaction force = R₁ x 5.5
= 5.5 R₁ upwards
Torque by weight of woman in opposite direction , downwards
= - 804 x ( 5.5 - 1.55 )
= - 3175.8
Torque by weight of the plank in opposite direction , downwards .
= - 27 x 5.5 / 2
= - 74.25
Torque by R₂ will be zero as it passes through the point about which torque is being taken .
Total torque
= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )
5.5 R₁ = 3250
R₁ = 590.9 N .
Answer:
KE = 2.03 J
Explanation:
After impact, the kinetic energy of the bullet+block will convert to potential energy
½mv² = mgh
v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s
conservation of momentum during the collision
0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)
u = 16.4481 m/s
KE = ½mv² = ½(0.015)16.4481² = 2.0290499...
KE = 2.03 J
Answer:
Frequency, f = 0.63 Hz
Period, T = 1.58 s
Speed of a wave, v = 1.34 m/s
Explanation:
The equation of a wave is given by :
...(1)
y is in mm
x is in meters
t is in seconds
The general equation of a wave is given by :
...(2)
(i) Compare equation (1) and (2) we get :
Since,
(ii) Period of wave is :
(iii) Speed of a wave,
