1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
expeople1 [14]
3 years ago
14

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

ce of the earth
Physics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

You might be interested in
the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight
djyliett [7]
Weight = (mass) x (gravity)

Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton
8 0
3 years ago
To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
Define impulse and momentum.<br>No spam​
Doss [256]

Impulse: a certain amount of force you apply for an amount of time.

Impulse: F*t where F= Force & t=time

Momentum: increasing forward motion.

A ball rolling down a slide gains momentum

p=mv where m=mass and v=velocity

Hope it helps!

~Just an emotional teen who listens to music

4 0
2 years ago
Learning Goal:To understand and be able to use the rules for determining allowable orbital angular momentum states.Several numbe
Free_Kalibri [48]

Answer:

Explanation:

1) for a given n value the l value can be from 0 to n-1

So if n= 5 it can take 0,1,2,3,4

i.e it can take 5 values

2)for an electron with l =3

it can be from -3 -2 -1 0 1 2 3

i.e it can take 7 values

3) n = 3 !!

l = 0 , 1 , 2

for l=0 , m = 0 total = 1

for l= 1 ,m = -1,0,1 total = 3

for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

total possible states = 9 * 2 = 18

Answer is 168

4)given l=3 and n=3

orbital quantum number cannot be equal to principal quantum number

its max value is l-1 only

5)L = sqrt(l(l+1))x h'

for it to be max l should be max

for n = 3 max l value is 2

therfore it is sqrt(2(2+1)) x h'

\sqrt(6) \times h'

this is the answer

5 0
3 years ago
The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn
AlexFokin [52]

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

1\ knot=0.514\ m/s

31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s

Power is given by

P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

5 0
3 years ago
Other questions:
  • in 1992 Maurizio damilano ,of italy walked 29752m in 2.00h( a) calcula damilano's average in m/s. suppose damilano slows down to
    11·1 answer
  • How many atoms are in Cu3(PO4)2
    9·2 answers
  • What is the magnitude of the net torque on the pulley about the axle?
    10·1 answer
  • While online last week, you saw the following advertisement:
    11·2 answers
  • The chewing muscle, the masseter, is one of the strongest in the human body. It is attached to the mandible (lower jawbone) as s
    15·1 answer
  • In a bag there are 18 pink jellybeans, 22 purple jellybeans, 10 orange jellybeans, and 20 red jellybeans what is the probability
    15·1 answer
  • A kilometer is a measure of an object’s
    13·1 answer
  • Suppose that the speed of a ball moving in a horizontal circle is increasing at a steady rate. Is this increase in speed produce
    15·2 answers
  • A space rover weighs less on Mars than it does on Earth. Which statement explains this difference? A. The gravitational constant
    10·1 answer
  • 21. A person standing 49.5m from the foot of a cliff claps his hands and hears an echo 0.3
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!