Answer:
92.81 psia.
Explanation:
The density of water by multiplying its specific gravity by the density of sea water.
SG = density of sea water/density of water
ρ = SG x ρw
1 kg/m3 = 62.4 lbm/ft^3
= 1.03 * 62.4
= 64.27lbm/ft^3.
The absolute pressure at 175 ft below sea level as this is the location of the submarine.
P = Patm +ρgh
= 14.7 + 64.27 * 32.2 * 175
Converting to pound force square inch,
= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )
= 14.7 + 78.11 psia
= 92.81 psia.
Yes!
I think there are two ways you could go with this answer:
1) Acceleration is the change in velocity over time, it can be negative or positive. If you have an object that is already moving forwards in a straight line and give it a constant negative acceleration, it will slow down and then start going in reverse.
2)Velocity is a vector, meaning it has both magnitude and direction. In the example above, the acceleration is due to a change in magnitude, or speed (from +ve to -ve) but not a change in direction. Something that has constant speed but is changing direction is also accelerating (like something that is orbiting). You could use the earth as an example, which is constantly accelerating due to moving in a circle around the sun. At any time in the year you can say that in half a year's time the earth's direction will be reversed.
To develop the problem it is necessary to apply the equations related to the moment of inertia.
The given values can be defined as,
![M = 1.0 kg](https://tex.z-dn.net/?f=M%20%3D%201.0%20kg)
![r = 0.5 m](https://tex.z-dn.net/?f=r%20%3D%200.5%20m)
![m = 10 g](https://tex.z-dn.net/?f=m%20%3D%2010%20g)
![I = 0.280 kg.m^2](https://tex.z-dn.net/?f=I%20%3D%200.280%20kg.m%5E2)
According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,
![I = I_{rim} + n * I_{spoke}](https://tex.z-dn.net/?f=I%20%3D%20I_%7Brim%7D%20%2B%20n%20%2A%20I_%7Bspoke%7D)
Where n is the number of spokes necessary to construct the wheel.
![I_{rim} = M*r^2 = 1.0 * 0.5^2](https://tex.z-dn.net/?f=I_%7Brim%7D%20%3D%20M%2Ar%5E2%20%3D%201.0%20%2A%200.5%5E2)
![I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2](https://tex.z-dn.net/?f=I_%7Bspoke%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%2A%20m%20%2A%20r%5E2%20%3D%20%5Cfrac%7B1%7D%7B3%7D%2A%2010%20%2A%2010%5E-3%20%2A%200.5%5E2)
Replacing the values at the general equation we have,
![0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )](https://tex.z-dn.net/?f=0.280%20%3D%201.0%20%2A%200.5%5E2%20%2B%20n%20%2A%20%281%2F3%20%2A%2010%20%2A%2010%5E-3%20%2A%200.5%5E2%20%29)
Solving for n,
![n = 36](https://tex.z-dn.net/?f=n%20%3D%2036)
Therefore the number of spokes necessary to construct the wheel is 36
PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then
![M_w= M + n*m](https://tex.z-dn.net/?f=M_w%3D%20M%20%2B%20n%2Am)
![M_w = 1.0 + 36* 10 * 10^{-3} Kg](https://tex.z-dn.net/?f=M_w%20%3D%201.0%20%2B%2036%2A%2010%20%2A%2010%5E%7B-3%7D%20Kg)
![M_w = 1.36 Kg](https://tex.z-dn.net/?f=M_w%20%3D%201.36%20Kg)
Therefore the mass of the wheel must be of 1.36Kg
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Answer:
Gravitational force, F = 1054.65 N
Explanation:
Given,
The accelerating speed, a = 15 m/s²
The mass of your body, m = 155 lbs
= 70.31 Kg
The gravitational force acting in a body is given by the relation
F = m x g
Where g is the acceleration due to gravity of the in which the velocity of the body changes its speed at a constant rate.
∴ a = g
Substituting the values in the above equation
F = 70.31 x 15
= 1054.65 N
Hence, the gravitational force acting on you, F = 1054.65 N