All categories

4 years ago

The work output of a machine is less than the work input what factor cause this

1 answer:

Ivan4 years ago

4 0

Total work energy on the input side is WE = Fs; where F is a force acting on a mass to push it s distance. This is the so-called work function. Let fs = we, which is the work energy (useful energy) attained as output when WE is input.

From the conservation of energy WE = Fs = fs - kNs = Total Output energy. Net force f = F - kN where kN is friction force acting against the pushing (input) force F. In the real world, there is always friction at some level. That is kN > 0 always. 

Thus Fs = (F - kN)s; kNs = the energy lost to friction where k is the friction coefficient and N is the normal force on the surface(s) where the friction is generated. By definition, efficiency = fs/Fs = useful work/work input. Clearly fs = Fs - kN < Fs . Thus efficiency = fs/Fs < 1.00, which means output fs < Fs the input whenever kN > 0, which in the real world it always is. 

The short answer is...output is always less than input because of friction and, sometimes, other losses like wind drag (which is a form of friction anyway).

You might be interested in

If a person weighs 600N on earth, how much does he weigh on the moon

Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0

4 years ago

A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m

victus00 [196]

Answer:

The acceleration of the car will be $a=9600m/sec^$

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion $s=ut+\frac{1}{2}at^2$

$200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2$

$a=9600m/sec^$

So the acceleration of the car will be $a=9600m/sec^$

6 0

4 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of

Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = $\frac{2L}{3}$

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center = 0.25 × wavelength

Distance = $0.25 x \frac{2L}{3}$

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = $0.25 x \frac{2 \times 39}{3}$

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

So, option C is correct.

7 0

3 years ago

A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs

Contact [7]

Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

4 0

4 years ago