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3 years ago

Two machines do thae same may have different power why

1 answer:

5 0

The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.

Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

Power = Work / time

P = W / t

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PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f

anzhelika [568]

Answer:

$\phi_i = BA$

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

$\phi = \vec B. \vec A$

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

$\phi = BAcos0 = BA$

6 0

3 years ago

How do you write a hypothesis

Mashcka [7]

It's and if, then statement!

5 0

3 years ago

Help me fast and i brainliest.

nexus9112 [7]

There are correlation and causation between the force of the finger and the movement of the books

8 0

3 years ago

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

4 0

3 years ago