Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
Answer:
915 Hz
Explanation:
The observed frequency from a sound source is given as
f₀ = f [(v + v₀)/(v+vₛ)]
where
f₀ = observed frequency of the sound by the observer = ?
f = actual frequency of the sound wave = 983 Hz
v = actual velocity of the sound waves = 343 m/s
vₛ = velocity of the source of the sound waves = 55.9 m/s
v₀ = velocity of the observer = 28.4 m/s
f₀ = 983 [(343+28.4)/(343+55.9)]
f₀ = 915.2 Hz = 915 Hz
Answer:
The velocity at the top of its path will be zero (0)
Explanation:
We can solve this problem or particular situation using the principle of energy conservation.
Which tells us that energy is transformed from kinetic energy to potential energy and vice versa. A reference point should be considered at which the potential energy is zero, and at this point the initial velocity of 40 [m/s] is printed to the ball.
![Ek=Ep\\where:\\Ek=kinetic energy [J]\\Ep=potencial energy [J]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5Cwhere%3A%5C%5CEk%3Dkinetic%20energy%20%5BJ%5D%5C%5CEp%3Dpotencial%20energy%20%5BJ%5D)
The potential energy is determined by:
![Ep=m*g*h\\where:\\m=mass of the ball[kg}\\g=gravity[m/s^2]\\h=heigth [m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3Dmass%20of%20the%20ball%5Bkg%7D%5C%5Cg%3Dgravity%5Bm%2Fs%5E2%5D%5C%5Ch%3Dheigth%20%5Bm%5D%5C%5C)
The kinetic energy is determined by:
![Ek=\frac{1}{2}*m*v_{0} ^{2} \\where\\v_{0} = initial velocity[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B0%7D%20%5E%7B2%7D%20%20%5C%5Cwhere%5C%5Cv_%7B0%7D%20%3D%20initial%20velocity%5Bm%2Fs%5D)
![Ek=Ep\\\frac{1}{2} *m*v_{0} ^{2} =m*9.81*h\\h=\frac{40^{2}}{2*9.81} \\h=81.5[m]](https://tex.z-dn.net/?f=Ek%3DEp%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7B0%7D%20%5E%7B2%7D%20%3Dm%2A9.81%2Ah%5C%5Ch%3D%5Cfrac%7B40%5E%7B2%7D%7D%7B2%2A9.81%7D%20%5C%5Ch%3D81.5%5Bm%5D)
This will be the maximum path but, its velocity at this point will be zero. Because now all the kinetic energy has been transformed in potential energy.
Answer:
The acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Explanation:
Given that,
Mass of the planet, m = 1/2M
Radius of the planet, r = 2R
Where M and R is the mass and radius of the Earth respectively.
The acceleration due to gravity of Earth, g = 10 m/s²
The acceleration due to gravity of Earth is given by the relation,
g = GM/R²
Similarly, the acceleration due to gravity of that planet is
gₐ = Gm/r²
where G is the Universal gravitational constant
On substituting the values in the above equation
gₐ = G (1/2 M)/4 R²
= GM/8R²
= 1/8 ( 10 m/s²)
= 1.25 m/s²
Hence, the acceleration due to gravity of that planet is, gₐ = 1.25 m/s²
Answer:
35.71 m
Explanation:
Potential energy is calculated using this formula:
- PE = mgh
- where m = mass (kg). g = gravitational acceleration on Earth (9.8 m/s²), h = height (m)
We are given 3 out of the 4 variables in this problem. We want to solve for h, the height of the cannon ball.
List out the known variables:
- PE = 14,000 J
- m = 40 kg
- g = 9.8 m/s²
- h = ? m
Substitute these values into the potential energy formula.
- 14,000 = (40)(9.8)h
- 14,000 = -392h
- h = 35.7142857143
The cannonball was 35.71 m high to have a potential energy of 14,000 J.