Ask question

Ask question

All categories

2 months ago

6

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

the escape speed ve of the rock from the planet. If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of _____ .

1 answer:

ValentinkaMS [3.4K]2 months ago

4 0

At great distances from the planet, if neglecting other gravitational influences, the rock will reach a speed close to $\sqrt{3} v_{e}$

Explanation:

To express the velocity that is very far from the planet alongside the escape velocity, we utilize the principles of energy conservation, allowing us to derive

The initial velocity of the rock, $v_{i}=2 v_{e}$ . Given that the radius is R, we can determine the escape velocity using the following formula:

$\frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}$

$v_{e}^{2}=\frac{2 G M}{R}$

$v_{e}=\sqrt{\frac{2 G M}{R}}$

Where M represents the mass of the planet and G represents the gravitational constant.

Based on the given conditions,

Surface potential energy can be articulated as, $U_{i}=-\frac{G M m}{R}$

As R approaches infinity when distanced from the planet, thus $v_{f}=0$

Then, the initialkinetic energy will be expressed as

$k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}$

Simultaneously, the finalkinetic energy can be described as

$k_{f}=\frac{1}{2} m v_{f}^{2}$

$v_{f}=\text { final velocity }$ Here,

Now, adding the potential and kinetic energies at the beginning and end gives the equation to determine the final velocity

expressed as

$U_{i}+k_{i}=k_{f}+v_{f}$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0$

$\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}$

$\frac{1}{2}$ By canceling 'm' common on both sides, we yield

$4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}$

Recognizing that $v_{e}=\sqrt{\frac{2 G M}{R}}$ , we can rewrite it as $\left(v_{e}\right)^{2}=\frac{2 G M}{R}$ , we arrive at

$4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}$

$v_{f}^{2}=3\left(v_{e}\right)^{2}$

Taking the squares out leads to

$v_{f}=\sqrt{3} v_{e}$

You might be interested in

When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the

Maru [3345]

Answer:

a

The value at a point inside is Zero

b

The electric field is $E = 2.7*10^{6} \ N/C$

Explanation:

We know from the problem that

The charge magnitude is $q = 3.0 \mu C = 3.0 *10^{-6} \ C$

The radius of the spherical ball is $r = 5.0 \ mm = 0.005 \ m$

According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero

On the outside, the electric field around the spherical ball is mathematically expressed as

$E = \frac{kq}{ a^2}$

Here a denotes a point outside the spherical ball with its value of $a = 10 \ cm = \frac{10}{100} = 0.1 \ m$

and k represents Coulomb's constant, valued at

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}$

=> $E = 2.7*10^{6} \ N/C$

5 0

1 month ago

A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or

Softa [3030]

Answer:

Baseball mass: $m_b=3.992*10^{14}kg$

Explanation:

Circumference of a baseball is calculated using 2πr = 23 cm

Thus, the radius comes out to be 3.66 cm, which equals $3.66*10^{-2}$ m

The mass density of the baseball matches that of a neutron or proton.

Proton mass = $10^{-27}$ kg

Proton diameter = $10^{-15}$ m

Proton radius = $5*10^{-16}$ m

Volume of the baseball is $\frac{4}{3} \pi r^3$

Now by substituting all values into the mass per unit volume equation for the baseball, we get:

$\frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}$

$\frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}$

$m_b=3.992*10^{14}kg$

Therefore, the baseball mass amounts to $m_b=3.992*10^{14}kg$

5 0

2 months ago

Read 2 more answers

If white wool (W) is dominant over black wool (w), can two sheep with white wool produce an offspring with black wool? Explain y

inna [3103]

The response is affirmative .
An organism exhibiting the dominant phenotype can have two potential genetic combinations (for a gene with two alleles):
It might be homozygous dominant (WW) or
it can be heterozygous dominant (Ww), which is also known as a carrier

For instance, two black sheep can produce offspring with white wool if both are heterozygous dominant. In this scenario, both parents could pass on the recessive allele, resulting in their offspring inheriting the phenotype of white wool characterized by the genotype ww.

6 0

2 months ago

Read 2 more answers

When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

kicyunya [3294]

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second = $\frac{20}{4}=5$

Spring's frequency of vibration is = $5 s^{-1}=5 Hz$

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring constant is 25 N/m.

3 0

3 months ago

Read 2 more answers