The plant that is closest to the sun is murcury. Then it is venus, then earth, and then mars. Then it is jupiter, then saturn, then uranus, then neptune.

5 0

3 years ago

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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

2 years ago

A flat horizontal surface with an area of 518 cm^2 is inside a uniform electric field of 1.33 x 10^4 N/C. The angle between the

GarryVolchara [31]

Answer: 576.48 N*m^2/C

Explanation: In order to calculate the electric flux through the any surface we have to take into account the scalar product between the electric field vector and the normal vector to the surface.

So we have:

ФE= E*A= 1.33 * 10^4*0.0518* cos (33.2°)= 576.48 N*m^2/C

3 0

2 years ago