Now write a short paragraph comparing "fast" to "speeding up quickly" and "slow" to "speeding up slowly".

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

3 years ago

Name the particle in the nucleus of an atom that gives the nucleus a positive charge

Crazy boy [7]

The answer is d...............

8 0

3 years ago

Read 2 more answers

The speed of a moving bullet can be deter-

Bad White [126]

Answer:

v = 381 m/s

Explanation:

Linear Speed

The linear speed of the bullet is calculated by the formula:

$\displaystyle v=\frac{x}{t}$

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

$\displaystyle \omega=\frac{\theta}{t}$

Where θ is the angular displacement and t is the time. Solving for t:

$\displaystyle t=\frac{\theta}{\omega}$

The angular displacement is θ=14°. Converting to radians:

$\theta=14*\pi/180=0.2443\ rad$

The angular speed is w=1436 rev/min. Converting to rad/s:

$\omega = 1436*2\pi/60=150.3776\ rad/s$

Thus the time is:

$\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}$

t = 0.0016 s

Thus the speed of the bullet is:

$\displaystyle v=\frac{0.61}{0.0016}$

v = 381 m/s

7 0

3 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.8 m/s rel

Svetach [21]

I'm assuming the question is what is the robin's speed relative to to the ground...

Create an equation that describes its relative motion.

rVg = rVa + aVg

Substitute values.

rVg = 12 m/s [N] + 6.8 m/s [E]

Use vector addition.

| rVg | = √ | rVa |² + | aVg |²

| rVg | = √ 144 m²/s² + 46.24 m²/s²

| rVg | = √ 190.24 m²/s²

| rVg | = 13.78 m/s

Find direction.

tanФ = aVg / rVa

tanФ = 6.8 m/s / 12 m/s

Ф = 29°

Therefore, the velocity of the robin relative to the ground is 14 m/s [N29°E]

4 0

3 years ago

Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m

Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0

3 years ago