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andrezito [222]

2 years ago

11

A hand crank generator is used to power a light bulb. The person turning the crank uses 22 Newtons of input force to turn the cr

ank 8 revolutions in 3 seconds. Each revolution of the crank equals 0.3 meters of distance that his hand moves. A multimeter connected to the light bulb records 0.08 amps of current and 9.04 volts.
1. Calculate the input power of the hand crank generator in Watts.

2. Calculate the output power of the hand crank generator in Watts.

3. Calculate the efficiency of the hand crank generator.

1 answer:

ikadub [295]2 years ago

8 0

Answer:

1. The input power is 17.6 Watts

2. The output power is 0.7232 Watts

3.The efficiency of the hand crank generator is approximately 4.109%

Explanation:

The given parameters are;

The force with which the crank generator is turned = 22 Newtons

The rate at which the crank is turned = 8 revolutions per 3 seconds

The distance the hand moves during each revolution = 0.3 meters

The current recorded by the multimeter = 0.08 amps

The voltage recorded by the multimeter = 9.04 volts

1. Power = The rate of doing work = Work done/(Time taken to do the work) = Force × Distance/Time

∴ The power input of the hand crank generator = 22 × 0.3 × 8/3 = 17.6 Watts

2. The power output to the bulb, P, is given by the formula for electrical power as follows;

Power = Current, I × Voltage V

∴ P = 0.08 × 9.04 = 0.7232 Watts

3. Efficiency in percentage = Output/Input × 100

Therefore, the efficiency of the hand crank generator = )Output power/(Input power)) × 100 = (0.7232/17.6) × 100 ≈ 4.109%

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A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting

sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

F dx = m v dv

We replace and integrate

-β ∫ x³ dx = m ∫ v dv

β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

x_max⁴ = 2 m /β v₀²

Let's look for the speed that the train can have for maximum compression

x_max = 20 cm = 0.20 m

v₀ =√(β/2m) x_max²

Let's calculate

v₀ = √(640 106/2 7.8 104) 0.20²

v₀ = 64.05 0.04

v₀ = 2,562 m / s

v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

v₀ = 9.2 km / h

5 0

2 years ago

A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz

nikitadnepr [17]

Answer:

25 - $\sqrt[4]{26.66*10^{-8} }$ mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2 $\pi$ f = 2 * $\pi$ * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter using the relation

$\frac{J}{c_{2} } = \frac{T}{t_{max} }$ = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

$\frac{J}{c_{2} }$ = $\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )$ = $\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ]$

therefore; 0.025^4 - $c^{4} _{1}$ = 0.050 / $\pi$ (7.8 *10^-6)

$c^{4} _{1}$ = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

$c_{1} = \sqrt[4]{26.66 * 10^{-8} }$ =

THE TUBE THICKNESS

$c_{2} - c_{1}$ = 25 - $\sqrt[4]{26.66*10^{-8} }$ mm

4 0

2 years ago

A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75 . The nozzle is supplied from an air reser

Flura [38]

Answer:

a. 4.279 MPa

b. 3.198 MPa to 4.279 MPa

c. 0.939 MPa

d. Below 3.198 MPa

Explanation:

From the given parameters

M $_{exit}$ = 1.75 MPa

M at 1.6 MPa gives A $_{exit}$ /A* = 1.2502

M at 1.8 MPa gives A $_{exit}$ /A* = 1.4390

Therefore, by interpolation, we have M $_{exit}$ = 1.75 MPa gives A

However, we shall use M $_{exit}$ = 1.75 MPa and A

Similarly,

P $_{exit}$ /P₀ = 0.1878

a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have

A $_{exit}$ /A* = 1.387. by interpolation M

Therefore P $_{exit}$ = P₀ × P

Which shows that the nozzle is choked for back pressures lower than 4.279 MPa

b) Where there is a normal shock at the exit of the nozzle, we have;

M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa

Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406

Where the normal shock occurs at the nozzle exit, we have

$P_b = 3.406\times 0.939 = 3.198 MPa$

Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as $P_b$ = 4.279 MPa

Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa

c) At M $_{exit}$ = 1.75 MPa and P

d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane

8 0

2 years ago

Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Yuki888 [10]

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

$60.1cm*\frac{1m}{100cm}=0.601m$

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

$V_{water}=\pi r^{2}h$

$V_{water}=\pi (\frac{0.601m}{2})^{2}*120m$

$V_{water}=113.28m^{3}$

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

$d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}$

$d_{water}=1000\frac{Kg}{m^{3}}$

Now, water density is given by the equation $d=\frac{m}{V}$ , where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

$m_{water}=d_{water}.V_{water}$

$m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}$

$m_{water}=113280Kg$

With the water mass we can find the weight of water:

$w_{water}=m_{water} *g$

$w_{water}=113280kg*9.8\frac{m}{s^{2}}$

$w_{water}=1110144N$

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

$w_{total}=w_{water}+w_{pipe}$

$w_{total}=1110144N+2500N$

$w_{total}=1112644N$

Converting this total weight to kN, we have:

$1112644N*\frac{0.001kN}{1N}=1113kN$

7 0

2 years ago

In which of the following branches of engineering is the practice not restricted?

fgiga [73]

Answer:

a) civil engineering.

Explanation:

Civil engineering is a professional engineering program that deals with the construction, design, and maintenance of all the natural and man-made environments including dams, buildings, railways, and roads.

Civil engineering is the branch of engineering that is the practice not restricted because civil engineer is not restricted to academic profession but practice in designing and construction can make someone a professional civil engineer.

Hence, the correct answer is "a)."

6 0

2 years ago

Read 2 more answers