80 foot-pounds of work is needed to move the sofa in Tyler's apartment. Which of the following statements is true?

What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

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

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

6 0

2 years ago

A student is collecting the gas given off from a plant in bright sunlight at a temperature of 27°c. The gas being collected is p

Brums [2.3K]

Answer: The gas being collected is probably Oxygen

Explanation:

Plants produce oxygen through a process known as photosynthesis by utilizing carbon monoxide, a by-product produced by humans during the process of breathing. Humans also breathe in the oxygen ( by-product of photosynthesis) produced by the plants. Therefore, humans and plants live in a symbiotic relationship.

Photosynthesis is the process plants use to synthesize food from carbon dioxide and water. Sunlight is used as a energy source. Photosynthesis releases oxygen as a byproduct.

At low temperatures, between 0 and 10 degrees Celsius – the enzymes that carry out photosynthesis do not work efficiently, and this decreases the photosynthetic rate.

At medium temperature above 10 degrees Celsius to below 40 degrees Celsius (e.g 27 degrees Celsius), the photosynthetic enzymes work at their optimum levels, so photosynthesis proceeds.

At a temperature above 40 degrees Celsius, the enzymes that carry out photosynthesis lose their shape and functionality, and the photosynthetic rate declines rapidly.

Equation for photosynthesis reaction:

6CO2 + 6H20 + (energy) = C6H12O6 + 6O2

4 0

1 year ago

What type of roadway has the highest number of hazards per mile?

Oliga [24]

The roadway with the highest number of hazards is city streets

4 0

1 year ago

The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n

qaws [65]

Felectric = q*E
 Ftranslational = m*a
 Felectric = Ftranslational
 q*E = m*a
 Solve for a
 a = q/m*E 
 Our sign convention is "up is positive"
 q = 1.6*10^-19 C
 m = 1.67*10^-27 kg
 E = -150 N/C (- because it is down and up is positive)
 a = -6,4*10^5 m/s^2 (downward)
 answer
a = -6,4*10^5 m/s^2 (downward)

3 0

2 years ago

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

2 years ago