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olga nikolaevna [1]
3 years ago
14

You do 1200J of work with gears.If the gears do 1000J of work, what is the efficiency of the gears?

Physics
1 answer:
slamgirl [31]3 years ago
8 0

Answer:

83.33%

Explanation:

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Define moment of momentum. at which condition is it's magnitude zero?​
ololo11 [35]

Let's start with the concept of momentum. What is it? Linear momentum in physics is mathematically written as a product of mass and velocity of an object. Now let us suppose a body of mass m is moving in an inertial frame of reference with velocity v. Consider the fact that no external force is acting on the system. The momentum of this body is given by mv, where m is the mass and v is its velocity. In case of simple real world problems not delving into the realms of relativity, mass is a conserved quantity and it cannot be zero. Hence the velocity of the body must be zero and hence the momentum.

However, photons are considered to have a rest mass zero.

However note the point carefully "rest mass". A body in motion cannot have mass to be zero.

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3 years ago
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3 years ago
The ____ of an object is a measure of the amount of matter it contains. on the other hand ____ is a measure of the gravitational
Phoenix [80]
<span>The Gravitational Force of an object is a measure of the amount of matter it contains. on the other hand __Matter__ is a measure of the gravitational force on an object. I hope it helps :)</span>
7 0
3 years ago
If a substance has a density of 2.7g/cm3 and a mass of 86.4g, what is its volume?
Gennadij [26K]

Answer:

32cm³

Explanation:

Given parameters:

  Density of substance  = 2.7g/cm³

   Mass of substance  = 86.4g

Unknown:

Volume of substance  = ?

Solution:

Density is the mass per unit volume of a substance.

    Density  = \frac{mass}{volume}

Since the unknown is volume we solve for it;

   mass  = density x volume

   86.4 = 2.7 x volume

    volume  = \frac{86.4}{2.7}   = 32cm³

7 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
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