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3 years ago

A light bulb that has more current running through it will...

Physics

1 answer:

RoseWind [281]3 years ago

4 0

C
It does not affect the brightess of a lightbulb

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What is the mass of 5 moles of gold

Serhud [2]

First of all, we know that one mole is equal to the atomic number of an element.

The atomic number of gold is <span>197.0g Au

And we need to find 5 moles.

5 * 197.0 g Au = </span><span>985.0g

Grams is used to measure mass.

Answer: </span>985.0g

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3 years ago

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HELPPPPPPP PLEASE!!! does anybody have the answer key to edge. physics lab report : mechanical equivalent to heat? one page of d

Karo-lina-s [1.5K]

Answer:

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case

Explanation:

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3 years ago

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14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

Zolol [24]

Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

5 0

3 years ago

What is the density of a roll of pennies containing 50 pennies?

ololo11 [35]

Answer:

Density is defined as:

Density = Mass/Volume

Now, density is an intensive property, this means that if you have 10 grams of a given material or 1000 grams of the same material, in both cases you will find the same density.

Then a roll of 50 pennies has the same density that a single penny.

The measures of a single penny are:

Mass = 2.5 g

Thickness = 1.52 mm

Radius = 9.525 mm

The coin is a cylinder, and the volume of a cylinder is:

V = pi*r^2*h

where:

pi = 3.14

r = radius = 9.525mm

h = thikness = 1.52mm

The volume is:

V = 3.14*(9.525mm)^2*1.52mm = 433.015 mm^3

The density will be:

D = 2.5g/433.015mm^3 = 0.00577 g/mm^3

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3 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

3 years ago