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3 years ago

Which planet is smaller than EarthÍs core?

1 answer:

3 0

The Earth's core is estimated to be bigger than both Mars and Mercury. However, I would say that Mercury is the safest choice. Mercury is also known to be smaller than Jupiter's Moon!

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Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

= - GMm/ (R+h)

Potential difference

= GMm/ R - GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0

3 years ago

You have a 100 ohm resistor. How

sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

$RT = \frac {R1*R2}{R1 + R2}$

Substituting into the formula, we have;

$75 = \frac {100*R2}{100 + R2}$

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

7500 + 75R2 = 100R2

7500 = 100R2 - 75R2

7500 = 25R2

R2 = 7500/25

R2 = 300 Ohms

4 0

3 years ago

A scientist bred fruit flies in two separate containers with different food sources for many generations. When she put the fruit

Nookie1986 [14]

Fruit flies prefer mates adapted to the same food source.

3 0

3 years ago

Read 2 more answers

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0

3 years ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

IrinaK [193]

Answer:

The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

We are given;

The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration is given by;

α = (ωₓ - ωₐ) / T₁

α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²

8 0

4 years ago