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KATRIN_1 [288]
3 years ago
10

Suppose the earth suddenly came to halt and ceased revolving around the sun. The gravitational force would then pull it directly

to the sun. What would be the earth's speed as it crashed?
Physics
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

613373.65233 m/s

Explanation:

M = Mass of Sun = 1.989\times 10^{30}\ kg

m = Mass of Earth

v = Velocity of Earth

r = Distance between Earth and Sun = 147.12\times 10^{9}\ m

r_e = Radius of Earth = 6.371\times 10^6\ m

r_s = Radius of Sun = 695.51\times 10^6\ m

In this system it is assumed that the potential and kinetic energies are conserved

\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s

The velocity of Earth would be 613373.65233 m/s

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The value of each resistor is equal to (Ps/Pp) - 2 Ohms.

<h3>How to determine the value of each resistor?</h3>

Let the numerical value of two unknown resistors be R₁ and R₂ respectively.

Based on the information provided, the total equivalent resistance of these two unknown resistors connected in series with a battery is given by:

Rt = R₁ + R₂

Also, power is given by:

P = I²R

Ps = I²(R₁ + R₂)     .....equation 1.

When the resistors are connected in parallel, we have:

Rt = (R₁R₂/R₁ + R₂)

Pp = I²(R₁ + R₂)     .....equation 2.

Dividing eqn. 1 by eqn. 2, we have:

Ps/Pp = (R₁ + R₂)²/R₁R₂

Ps/Pp = [R₁(1 + R₂/R₁)²]/R₁R₂

Let x = R₂/R₁;

So, the power becomes;

Ps/Pp = x(1 + x)²

xPs/Pp = x² + 2x + 1

Ppx² + 2xPp + Pp = xPs

Ppx² + 2xPp + Pp - xPs = 0

x² + [(2Pp - Ps)/Pp]x + 1 = 0

Next, we would solve for x by using the quadratic formula:

x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{-(\frac{2P_p -P_s)}{P_p} )\; \pm \;\sqrt{(\frac{2P_p -P_s)}{P_p} )^2 - 4 }}{2}

x = (Ps/Pp) - 2

Therefore, x = R₂/R₁ = (Ps/Pp) - 2 Ohms.

Read more on resistance in parallel here: brainly.com/question/23282393

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The ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

<h3>What is the rotational kinetic energy of a sphere?</h3>

The rotational kinetic energy of the sphere is directly proportional to the square of angular acceleration.

K = \dfrac 12 I\omega ^2

Where,

I = rotational inertia of sphere = 2/5MR²

Where M is the mass of the sphere and R is the radius of the sphere,

ω = angular acceleration of sphere = V/R

Where V is the speed of the sphere.

So,

K = 1/2 × 2/5MR² × V²/R²

K = MV²/5

The translational kinetic energy of the sphere,

K' = \dfrac 12 MV^2

The total kinetic energy of the sphere,

K_t = K + K'

So,

K_t= \dfrac {MV^2}5 +\dfrac { MV^2}2\\\\K_t= \dfrac {7MV^2}{10}

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\dfrac KK_t = \dfrac {\dfrac {MV^2}{5} }{ \dfrac {7MV^2}{10}}\\\\\dfrac KK_t = \dfrac 15 \times \dfrac { 10}7\\\\\dfrac KK_t= \dfrac 27

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Answer:

Nearest, the revolutions per minute will be 29.

Explanation:

Given that,

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Where, n = number of revolutions in one second

We need to calculate the revolutions in one second

Using formula of centripetal acceleration

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Put the value of a and ω

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Put the value into the formula

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Using value for the revolutions per minute

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