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Gelneren [198K]
3 years ago
5

The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitu

de of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2
Physics
1 answer:
Trava [24]3 years ago
8 0

Answer:

the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)

Explanation:

according to newton's law of universal gravitation ( we will neglect relativistic effects)

F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet

if we assume that the planet has a spherical shape,  the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,

since M= ρ* V = ρ* 4/3πR³ , ρ= density

F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR

from Newton's second law

F= m*g = G*ρ*m* 4/3πR

thus

g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R

g ∝ R

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The electron gun in a television tube accelerates electrons (mass = 9.11x10^-31 kg, charge = 1.6 x 10^-19C) from rest to 3 x10^7
stich3 [128]

Answer:

Electric field acting on the electron is  127500 N/C.

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Charge on electron, q=1.6\times 10^{-19}\ C

Initial speed of electron, u = 0

Final speed of electron, v=3\times 10^7\ m/s

Distance covered, s = 2 cm = 0.02 m

We need to find the electric field required. Firstly, we will find the acceleration of the electron from third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

F = ma

F=9.1\times 10^{-31}\times 2.25\times 10^{16}

F=2.04\times 10^{-14}\ N

Electric force is given by :

F = q E, E = electric field

E=\dfrac{F}{q}

E=\dfrac{2.04\times 10^{-14}}{1.6\times 10^{-19}}

E = 127500 N/C

So, the electric field is 127500 N/C. Hence, this is the required solution.

8 0
3 years ago
If a 25 kg object is moving at a velocity of 10 m/s, the object has<br> energy. Calculate it.
Paul [167]

Answer: Your answer is 1250J

Explanation:

K E = 1/2 m v 2

The mass is  

m = 25 k g

The velocity is  v = 10 m s − 1

So,

K E = 1 /2 x25 x 10 2^2= 1250 J

pls mark brainiest answer  

4 0
2 years ago
Physics question, please help?
Ludmilka [50]

0.4823 m/s

The initial velocity u1 of the ball=0

From the law of conservation of linear momentum.

m1u1+m2u2=m1v1+m2v2

(160×0)+(170×u1)=(160×0.3)+(170×0.2)

u1=0.4823m/s

6 0
3 years ago
HELP ASAP!! 15 POINTS!!
aksik [14]

Answer:

Its probably D (the sound energy)

4 0
3 years ago
Two technicians are discussing electronic leveling systems. Technician A says that a weight should be placed in the vehicle as p
Ronch [10]

Answer:

C. Both A and B

Explanation:

<em>Electronic leveling systems: </em>These are automatic lifting jacks use for raising  wheels and vehicle when parked or when work is to be done on the vehicle engine underneath or on wheels. unlike the manual jacks it is very quick and efficient.

Both technician right, for Technician A, a weight may be use during the procedure for balancing and leveling when raising, and for Technician B, the sensor in the electronic leveling systems is adjustable to adjust the height so as to maintain balancing and levelling.

8 0
3 years ago
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