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3 years ago

What is an event horizon? Does our Sun have an event horizon around it?

Physics

1 answer:

STatiana [176]3 years ago

6 0

Answer:

The event horizon of a black hole can be thought of either as the place around the black hole where the speed you need to escape becomes greater than the speed of light or as the place where the warping of spacetime around a collapsed star becomes so great that all straight lines pointing outward actually become curved paths bringing you back in. Only black holes have event horizons, so our Sun, which is a star in happy main-sequence equilibrium, cannot have one.

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The atlas stone’s is a strong man competition where athletes have to load 5 stones of masses 100kg, 120kg, 140kg, 160kg and 180k

lyudmila [28]

Answer: 6250 joules

Explanation:

The work needed to lift an object of mass M by a height H is equal to:

w = M*g*H

where h = 10m/s^2

then the total work that he did is equal to the sum of the work for every stone:

W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)

= (100kg + 120kg + 140kg + 160kg + 180kg)*g*H

= (500kg)*g*H

and now we can repalce g by 10m/s^2 and H by 125cm

But you can notice that we have two different units of distance, so knowing that 100cm = 1m

we can write H = 125cm = (125/100) m = 1.25 m

Then we have:

H = 500kg*10m/s^2*1.25m = 6250 J

3 0

3 years ago

A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125

frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0

3 years ago

3 facts about the concept or term of tectonic plates

Agata [3.3K]

1) Plates movement happens very slowly.
2) The earth's lithosphere is divided into pieces
3) they are caused by convection currents in the asthenosphere (upper mantle)

8 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago

Starting from rest, a crate of mass m is pushed up a frictionless slope of angle theta by a horizontal force of magnitude F. Use

ludmilkaskok [199]

Answer:

v = sqrt[2*(F*h*cot(theta)-mgh)/m]

Explanation:

Work = KE + Ug

F*r=1/2mv^2+mgh

1/2mv^2=F*r-mgh

v=sqrt[2(F*r-mgh)/m]

r=h/tan(theta)=h*cot(theta)

6 0

3 years ago