Answer:
25.06s
Explanation:
Remaining part of the question.
(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)
Solution:
F = 60N
r = 90cm = 0.9m
M = 8200kg
Moment of inertia for a sphere (I) = ⅖mr²
I = ⅖ * m * r²
I = ⅖ * 8200 * (0.9)²
I = 0.4 * 8200 * 0.81
I = 2656.8 kgm²
Torque (T) = Iα
but T = Fr
Equating both equations,
Iα = Fr
α = Fr / I
α = (60 * 0.9) / 2656.8
α = 0.020rad/s²
The time it will take her to rotate the sphere,
Θ = w₀t + ½αt²
Angular displacement for one revolution is 2Π rads..
θ = 2π rads
2π = 0 + ½ * 0.02 * t²
(w₀ is equal to zero since sphere is at rest)
2π = ½ * 0.02 * t²
6.284 = 0.01 t²
t² =6.284 / 0.01
t² = 628.4
t = √(628.4)
t = 25.06s
Answer:
2.72×10^-7
Explanation:
velocity = frequency × wavelength
2.05×10^8=7.55×10^14 wavelength
wavelength = 2.05×10^8/7.55×10^14
wavelength = 2.72×10^-7
Answer:
The horizontal range will be
Explanation:
We have given initial speed of the shell u =
Angle of projection = 51°
Acceleration due to gravity
We have to find maximum range
Horizontal range in projectile motion is given by
So the horizontal range will be
F₁ = c / d²
F₂ = c / (3d)²
F₁/F₂ = 3² = 9
F₂ = 1/9 F₁