All categories

2 years ago

Derive that efficiency is the ratio of output and input work.

Physics

2 answers:

Stels [109]2 years ago

8 0

Answer:

The efficiency of a simple machine is defined as the ratio of useful work done by a machine (output work) to the total work put into the machine (input work). For ideal or perfect machine, work output is equal to the work input.

Explanation:

Hope it Helps!!!

spayn [35]2 years ago

4 0

Answer:

efficiency of a simple machine is defined as the ratio of useful work done by a machine (output work) to the total work put into the machine (input work). For ideal or perfect machine, work output is equal to the work input.

Explanation:

hope it will help you

You might be interested in

A lightning bolt may carry a current of 1.00 104 A for a short time. What is the resulting magnetic field 160 m from the bolt? S

ZanzabumX [31]

Answer:

Result magnetic field=

$1.25*10^-^5T$

Explanation:

We are given:

$I = 1.00*10^4A$

r = 160m

Let us say the lightning bolt is a long straight conductor.

Therefore we will use the formula:

B = Uo•I / 2πr

Substituting figures in the equation, we have:

$B = [(4π*10^-^7 T•m/A)(1.00*10^4)] / (2π * 160)$ ;

$= 1.25 * 10^-^5 T$ ;

$Therefore, the result magnetic field at 160m = 1.25 * 10^-^5$

4 0

3 years ago

If lebron james has a vertical leap of 1.29m, what is his takeoff speed?

BlackZzzverrR [31]

v=u+at 
0=5.0283−9.8×t
t=0.5130sec
Total time = 2t = 1.0261 sec

3 0

3 years ago

Read 2 more answers

Can you help me with this pls

devlian [24]

Answer:

Can you help me with this pls

4 0

3 years ago

A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$ (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

5 0

3 years ago

Read 2 more answers

A high-speed ""bullet"" train accelerates and decelerates at the rate 4 ft/s 2 . Its maximum cruising speed is 90 mi/hr. (a) Wha

NemiM [27]

Answer:

23 mi

Explanation:

Let us first record all the values we have

acceleration = 4 ft/s

max speed = 90 mi/hr * 1 hr/3600s * 5280 ft/1 mi = 132 ft/s

We can use the equation of linear motion since acceleration is constant

distance covered = distance during acceleration + distance during costant

velocity

$s = \frac{v_{f} ^{2} - v_{i} ^{2} }{2a} + v_{i}t\\\\s = \frac{132 ^{2} - 0^{2} }{2*4} + 132*900\\s = 120 978ft$

distance in miles = 120 978 ft * 1 mi/5280 ft = 22.9125 mi ≈ 23 mi

4 0

3 years ago

Derive that efficiency is the ratio of output and input work.​

Derive that efficiency is the ratio of output and input work.