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3 years ago

A temperature of 20°C is equivalent to approximately

Physics

2 answers:

igomit [66]3 years ago

8 0

Its the answer D. 68°F

algol133 years ago

7 0

A temperature of 20°C is equivalent to approximately

A. 32°F.

B. –6°F.

C. 136°F.

D. 68°F.

The correct answer for this question is D. 68°F

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One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R

Drupady [299]

(a) 120.8 m/s^2

The gravitational acceleration at a generic distance r from the centre of the planet is

$g=\frac{GM'}{r^2}$

where

G is the gravitational constant

M' is the mass enclosed by the spherical surface of radius r

r is the distance from the centre

For this part of the problem,

$r=R=1.17\cdot 10^6 m$

so the mass enclosed is just the mass of the core:

$M'=M=2.48\cdot 10^{24}kg$

So the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(2.48\cdot 10^{24}kg)}{(1.17\cdot 10^6 m)^2}=120.8 m/s^2$

(b) 67.1 m/s^2

In this part of the problem,

$r=3R=3(1.17\cdot 10^6 m)=3.51\cdot 10^6 m$

and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so

$M'=M+4M=5M=5(2.48\cdot 10^{24}kg)=1.24\cdot 10^{25}kg$

so the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(1.24\cdot 10^{25}kg)}{(3.51\cdot 10^6 m)^2}=67.1 m/s^2$

8 0

4 years ago

The weight of a person on Earth is 67.0 N. If the gravitational acceleration on the moon is 1.62 m/s2, calculate the person’s we

Rzqust [24]

Answer:

$\large{ \boxed{ \tt{b) \: 11.0 \: N}}}$

- Please see the attached picture for full solution!:)

--------------- HappY LearninG <3 ---------

5 0

2 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

Dafna1 [17]

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

Critics of the electoral college argue that the system can result in the selection of —

Romashka [77]

Answer:

President who did not win the popular vote

Explanation:

Critics of the Electoral college argue that the system can result in a selection of what?

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3 years ago

HELP HELP HELP IM DYING photosynthesis leafs

emmasim [6.3K]

The difference between A,B,C,D are that the yields go up a with the carbon dioxide for EF and G the temp goes up but not the carbon dioxide

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3 years ago