A satellite with a mass of 5.60 E5 kg is orbiting the Earth in a circular path at 7.037E 6 meters from

What is the primary visible color of an emission nebula?

zzz [600]

In emission nebulae, there are interstellar clouds of hydrogen, which glow red because of the intense radiation of hot stars inside the nebula

7 0

3 years ago

A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force e

WINSTONCH [101]

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

7 0

3 years ago

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Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th

satela [25.4K]

The tank has a volume of $\dfrac\pi3R^2H$ , where $H=6\,\rm m$ is its height and $R=\dfrac d2=2\,\rm m$ is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

$\dfrac26=\dfrac rh\implies r=\dfrac h3$

The volume of water in the tank at any given time is

$V=\dfrac\pi3r^2h$

and can be expressed as a function of the water level alone:

$V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3$

Implicity differentiating both sides with respect to time $t$ gives

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}$

We're told the water level rises at a rate of $\dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min}$ at the time when the water level is $h=2\,\mathrm m=200\,\mathrm{cm}$ , so the net change in the volume of water $\dfrac{\mathrm dV}{\mathrm dt}$ can be computed:

$\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}$