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A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer 1

Answer:

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

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A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Answers

The speed when all the fuel has been exhausted is 2415m/s

According to this question, the following information was given:

Exhaust velocity of fuel, V(e)= 1500 m/s
Initial speed of rocket, V₁ = 0 m/s
Final speed of rocket, V₂ = ?
Fuel weight = 80% of total weight

Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

m1 = initial mass
m2 = final mass without repellant

m2 = m1 - 80%m1

m2 = m1 - 0.8m1

m2 = 0.2m1

∆V = V2 - V1

Hence;

V2 - 0 = 1500 × ln (m1/0.2m1)

V2 = 1500 ln(1/0.2)

V2 = 1500 × 1.609

V2 = 2415m/s.

Therefore, the speed when all the fuel has been exhausted is 2415m/s.

Learn more at: brainly.com/question/19531823?referrer=searchResults

Answer:

$v_2 =2414\ m/s$

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using Tsiolkovsky rocket equation

$\Delta v = v_e ln((m_1)/(m_2))$

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the initial total mass.

m₂ is the is the final total mass without propellant.

m₂ = m₁ - 0.8 m₁

m₂ = 0.2 m₁

$v_2-v_1 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 = 1500* ln((m_1)/(0.2 m_1))$

$v_2 =2414\ m/s$

When all the fuel is exhausted speed of the fuel is equal to $v_2 =2414\ m/s$

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?

Answers

Hi, thank you for posting your question here at Brainly.

To solve this problem, we use Coulomb's Law:

F = kQ1Q2/d^2, where k = 9x10^9

Q1 = 3.5 uC
Q2 = -3.5 uC
Q3 = 4.0 uC

But first, we find the distance between Q1 and Q3 and between Q2 and Q3.

d between Q1 and Q2:
d = sqrt[(0-0.4)^2+(0.3-0)^2]
d = 0.5 m

d between Q1 and Q3:
d = sqrt[(0-0.4)^2+(-0.3-0)^2]
d = 0.5 m

Through force balance, F between Q2 and Q3 - F between Q1 and Q3:

$F_(net) = ((9x 10^(9))(-3.5)(4) )/( 0.5^(2) ) -((9x 10^(9))(3.5)(4) )/( 0.5^(2) )=-1.008* 10^(12)$

Thus, the net force is -1 x 10^-12 C

Final answer:

The total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC is zero. This is because the forces due to each of these charges on the third charge are equal in magnitude but opposite in direction, hence they cancel each other completely.

Explanation:

The question asks for the magnitude and direction of the total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC. This is related to Coulomb's Law, which describes the force between charged objects. Specifically, Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1*q2) and inversely proportional to the square of the distance (r) between them. It also depends on the permittivity of free space (ε₀).

First, you would determine the force between each of the point charges and the third charge separately, and then superpose these forces to find the total force. The force in each case can be calculated using the equation F = k*|q1*q2|/r², where k is Coulomb's constant (8.99 * 10^9 N.m²/C²). You would need to make sure you take into account the signs of the charges when deciding the directions of the forces and when superposing the separate forces.

Assume upwards to be the positive direction. The 3.5 uC charge forces and -3.5 uC charge forces on the 4 uC charge would be opposite in direction (one downwards and one upwards) and identical in magnitude. Therefore, they will cancel each other out, and hence, the total electric force on the third charge (4 uC) will be zero.

Learn more about Total Electric Force here:

brainly.com/question/30448827

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If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change?a. 4
b. 2/pi
c. 2
d. (2)^1/2

Answers

Answer:

the answers the correct one is a 4

Explanation:

The centripetal acceleration is by

a = v² / R

angular and linear velocities are related

v = w R

let's substitute

a = w² R

for initial condition

a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

a = (2w₀)² R

a = 4 (w₀² R)

a = 4 a₀

when reviewing the answers the correct one is a

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

Answers

Here is the full question

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi ((D)/(2))^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= (A)/(N) =(\pi ((D)/(2))^2 )/(10, 000)$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= (d)/(\pi)$

$r_e = \sqrt{(d)/(\pi) }$

replacing d = $(\pi ((D)/(2))^2 )/(10, 000)$ in the equation above; we have:

$r_e = \sqrt{((\pi ((D)/(2))^2 )/(10, 000))/(\pi) }$

$r_e = \sqrt{(((D)/(2))^2 )/(10, 000)}$

$r_e = \sqrt{(((100,000)/(2))^2 )/(10, 000)}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

$m=0.0001 kg$ is the mass of the honeybee

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting into the equation, we find:

$W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N$

weight = mg
here m = 0.000100 g = 9.80
hence weight = 0.00980 kgm/s2

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Answers

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $(1)/(f)$

where

f = focal length

Thus

f = $(1)/(P)$

f = $(1)/(2)$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$(1)/(f) = (1)/(u) + (1)/(u')$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$(1)/(u') = (1)/(f) - (1)/(u)$

$(1)/(u') = (1)/(0.5) - (1)/(0.25)$

$(1)/(u') = (1)/(f) - (1)/(u)$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

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