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3 years ago

"two planets have the same mass, but planet a has 3 times the radius of planet

1 answer:

4 0

The gravitational acceleration on the surface of planet A is:
$g_A = \frac{GM_A}{r_A^2}$
where G is the gravitational constant, $M_A$ is the mass of planet A and $r_A$ its radius.

Similarly, the gravitational acceleration on the surface of planet B is:
$g_B = \frac{GM_B}{r_B^2}$

The ratio between the gravitational acceleration on planet A and B becomes:
$\frac{g_A}{g_B}= \frac{GM_A / r_A^2}{GM_B/r_B^2} = \frac{M_A r_B^2}{M_B r_A^2}$

The problem says that the two masses are equal: $M_A = M_B$ while planet A has 3 times the radius of planet B: $r_A = 3 r_B$ . Substituting into the ratio, we get:
$\frac{g_A}{g_B} = \frac{M_B r_B^2}{M_B (3 r_B)^2} = \frac{1}{9}$

so, gravity on planet B is 9 times stronger than planet A.

You might be interested in

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu

anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = $\frac{3Q}{4\pi R^{3} }$

So, $Q_{1} = \frac{Qr^{3} }{R^{3} }$

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0

3 years ago

Someone please help me with these questions! (The ones in the picture) Please I am super confused!

lozanna [386]

Answer:

c-d

Explanation:

3 0

2 years ago

The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from

uysha [10]

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

$U_{0}=Fx$ ....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

$F_{r}=12F$

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

Using formula for kinetic energy

$U=F_{r}x$

Put the value of $F_{r}$

$U=12Fx$

Now, from equation (I)

$U=12U_{0}$

Hence, The kinetic energy of the particle will be 12U₀.

7 0

3 years ago

PLEASE ANSWER, I NEED HELP

Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is $1.56\cdot 10^{-3} N$

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 47 kg$ is the mass of Ellen

$m_2 = 7.35\cdot 10^{22} kg$ is the mass of the moon

$r=3.84\cdot 10^8 m$ is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

$F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N$

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

$a=\frac{F}{m}$

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0

3 years ago

A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s

insens350 [35]

Answer:

24.531 m