Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :
v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.
Answer:
part a : <em>The dry unit weight is 0.0616 </em>
<em />
part b : <em>The void ratio is 0.77</em>
part c : <em>Degree of Saturation is 0.43</em>
part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>
Explanation:
Part a
Dry Unit Weight
The dry unit weight is given as
Here
is the dry unit weight which is to be calculated- γ is the bulk unit weight given as
- w is the moisture content in percentage, given as 12%
Substituting values
<em>The dry unit weight is 0.0616 </em><em />
Part b
Void Ratio
The void ratio is given as
Here
- e is the void ratio which is to be calculated
- is the dry unit weight which is calculated in part a
is the water unit weight which is 62.4 
or 0.04
- G is the specific gravity which is given as 2.72
Substituting values
<em>The void ratio is 0.77</em>
Part c
Degree of Saturation
Degree of Saturation is given as
Here
- e is the void ratio which is calculated in part b
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Substituting values
<em>Degree of Saturation is 0.43</em>
Part d
Additional Water needed
For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as
Here
is the zero air unit weight which is to be calculated
- is the water unit weight which is 62.4 or 0.04
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Now as the volume is known, the the overall weight is given as
As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.
Answer:
Technician A is right. The situation will happens even with only two bulbs in series
Explanation:
We must take into account that
1.- All electric device need its nominal voltage to operate
2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)
3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).
4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.
For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage
Answer:
Absolute error=0.006
Percentage Relative error=0.6
Explanation:
The resistors have resistance of 24 ohm and 8 ohm.
The change in resistance is 0.5 and 0.3 ohm respectively.
Relative error for parallel combination of resistors is
= dR/R²
= dR1/R1² + dR2/R2²
= 0.5/(24)² + 0.3/(8)²
= 0.5/ 24*24 + 0.3/ 64
= 0.5/576+0.3/ 64
= 32 + 172.8/ 36,864
=204.8/ 36,864=0.0055
=0.006
Percentage error =Relative error *100
= 0.006* 100 = 0.6
Answer:
88.3
Explanation:
Emf in a rotating coil is given by rate of change of flux:
E= dФ/dt=(NABcos∅)/ dt
N: number of turns in the coil= 80
A: area of the coil= 0.25×0.40= 0.1
B: magnetic field strength= 1.1
Ф: angle of rotation= 90- 37= 53
dt= 0.06s
E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V
