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zlopas [31]
1 year ago
14

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-s

ized 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 0.017 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)
Physics
2 answers:
Elodia [21]1 year ago
7 0

Answer:

Velocity will be v=3.266\times 10^5m/sec

Explanation:

We have given mass of the star M=2\times 10^{30}kg[/tex}Radius of the star [tex]R=5\times 10^{3}m

Gravitational constant G=6.67\times 10^{-11}Nm^2/kg^2

We know that acceleration is given by a=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times 2\times 10^{30}}{(5\times 10^3)^2}=5.33\times 10^{12}m/sec^2

Displacement is given as s = 0.017 m

From third equation of motion

v^2=u^2+2as

As initial velocity u = 0 m/sec

So v^2=0^2+2\times 5.33\times 10^{12}\times 0.017

v=3.266\times 10^5m/sec

laiz [17]1 year ago
5 0

Answer:

30298514.82 m/s

Explanation:

M = Mass of star = 2×10³ kg

r = Radius of star = 5×10³ m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

a=G\frac{M}{r^2}\\\Rightarrow a=6.67\times 10^{-11}\frac{2\times 10^{30}}{5\times 10^3}\\\Rightarrow a=2.7\times 10^{16}\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 2.7\times 10^{16}\times 0.017+0^2}\\\Rightarrow v=30298514.82\ m/s

The object would be moving at a velocity of 30298514.82 m/s

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This is a free fall and in free fall we use this formula:

d = (1 ÷ 2) × g × t², where d is the distance, g is the gravitational acceleration and t is the time.

In our case,

We are already given the moon's gravitational acceleration and we are going to substitute it with g. Let's leave the unknown alone, which is t.

t = √(2d ÷ g)

If we perform the formula, t is found to be √(2d ÷ g) = √(2 × 1.2 ÷ 1.62) ≅ 1.217 s

I am sorry for my bad English and if there is anything that you do not understand please let me know.

7 0
2 years ago
A solid spherical insulator has radius r = 2.5 cm, and carries a total positive charge q = 8 × 10-10 c distributed uniformly thr
ira [324]
At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m.  So substituting and clearing, we have that the magnitude of the electric field will be:  E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.
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An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
Natali [406]

Answer:

1.38*10^18 kg

Explanation:

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F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
2 years ago
Gravitational potential energy is often released by burning substances. true or false
jolli1 [7]
Gravitational potential energy is caused when an object is resting above the ground. It is released when the object is falling, not by burning substances.
4 0
1 year ago
Read 2 more answers
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
1 year ago
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