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malfutka [58]
3 years ago
8

Write two differences between practical machine and ideal machine​

Physics
1 answer:
slava [35]3 years ago
6 0

Answer:

Ideal machines are said to have parts which are weightless , frictionless and strings if any are inextensible. Practical machines have parts which have weights, friction and it causes some loss of work done. So these are the differences between Ideal Machine and Practical Machines.

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"What is the overall length limitation of a UTP cable run from the telecommunications closet to a networking device in the work
Goryan [66]

Answer:

90-100meters

Explanation:

The overall length limitation of a UTP cable is 90-100meters. Once this limitation is reached, a repeater is employed to transfer data.

4 0
4 years ago
A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg
Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0
3 years ago
Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.
valentina_108 [34]
The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics
5 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works
steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

  • 1km=10³m
  • 1m=10^{-3}km

Mercury

  • 58000m

Ven us

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Earth

  • 150000m

Mars

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Jupiter

  • 778000m

Saturn

  • 1430000m

Uranus

  • 2870000m

Neptune

  • 4500000m

7 0
2 years ago
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