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3 years ago

In Physics what type of distance is used in calculating work?

1 answer:

7 0

Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton. meter (N. ... One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m

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What is different about the colors of the spectrum?

ankoles [38]

The human eye is not capable of seeing radiation wavelengths outside the visible Spectrum, i hope this answers your question :)

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3 years ago

A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around

jolli1 [7]

Answer:

No, the farmer is not able to move the mule.

Explanation:

Mass =100 kg

Force=F=800 N

The coefficient between the mule and the ground= $\mu_s=0.8$

$\mu_k=0.5$

Static friction force,f= $\mu N$

Normal force=N=mg

Static friction force,f= $\mu_s mg=0.8\times 100\times 9.8=784 N$

Using $g=9.8m/s^2$

F<f

Static friction force is greater than applied force.

Therefore , the farmer is not able to move the mule.

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3 years ago

What is dark energy and what evidence do astronomers have that it is an important component of the universe?

PSYCHO15rus [73]

Answer:

The existance of dark energy was suggested to help explain measurements, using Type Ia supernovae as distance indicators, that the expansion of the universe is speeding up and its acceleration requires a source of energy. Scientists suggested that It may be a new form of energy for which there is not yet a theoretical explanation. quantum mechanics predict that it may be the vacuum energy associated with "empty" space itself.

8 0

3 years ago

5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0

3 years ago

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

3 years ago