B
Think of inertia of getting into a car accident without a seat belt although the car stops you will not you would likely fly out the window
Answer:
The velocity of the truck after the collision is 20.93 m/s
Explanation:
It is given that,
Mass of car, m₁ = 1200 kg
Initial velocity of the car, 
Mass of truck, m₂ = 9000 kg
Initial velocity of the truck, 
After the collision, velocity of the car, 
Let 
is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.
So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.
Answer:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
Answer: 7022.2kg/m³, yes, I was cheated
Explanation:
Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;
Density = Mass/Volume
Note that the unit of both mass and volume must be standard unit.
Given mass = 0.0158kg
Dimension of the metal = 5mm×15mm×30mm
Note that 1mm = 0.001m
The volume of the metal will be
0.005×0.015×0.03
= 0.00000225m³
Density = 0.0158/0.00000225
Average density of the metal = 7022.2kg/m³
Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.
Answer:
The maximum transverse speed of the bead is 0.4 m/s
Explanation:
As we know that the Amplitude of the travelling wave is
A = 3.65 mm
Now the speed of the travelling wave is
now we know that distance of first antinode from one end is 27.5 cm
so length of the loop of the standing wave is given as
now we have
now we have
now at x = 13.8 cm
now we have
now maximum speed is given as
