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3 years ago

8

a boat sailing against the current experiences an acceleration of -11 m/s^2 if the boats initial velocity is 44 m/s upstream, ho

w long until it comes to a stop?

1 answer:

valentinak56 [21]3 years ago

3 0

Answer:4s

Explanation:

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Fish in the aquarium looking much bigger than they are.

Alexandra [31]

b. Refraction

Explanation:

Fish in the aquarium are looking much bigger than they are due to refraction.

Refraction causes the path of light to bend and the velocity and wavelength attenuated.

This phenomenon occurs when light travels between two materials of different densities.

Air has a refractive index of 1
Water has a refractive index of 1.33
When light form water comes to air, the their angle changes.
This causes the magnification of the fish.

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Refraction brainly.com/question/12370040

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4 years ago

What is the weight of an object with a force of 35 pounds ?

Inga [223]

Answer:

156 Newton I think if it's wrong correct me

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3 years ago

Complete the paragraph to describe the relationship between kinetic energy and braking distance. Use . A car moves at a speed of

Mariana [72]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

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4 years ago

Read 2 more answers

If the same satellite orbited at the same velocity around a planet with three times the force of gravity, the orbit radius would

belka [17]

Answer:

the orbit radius would be divided by 3

Explanation:

In order to orbit the planet, the centripetal force of satellite must balance the gravitational force between planet and satellite.

centripetal Force = Gravitational Force

(Ms)(V)²/r = (G)(Ms)(Mₓ)/r²

V² = GMₓ/r

where,

V = velocity of satellite

G = Gravitational Constant

Mₓ = Mass of Planet

r = orbit radius

but, G = gr²/Mₓ

Therefore,

V² = (Mₓ/r)(gr²/Mₓ)

V = √gr --------------------- equation (1)

where,

g = force of gravity

Now, for change in force of gravity:

V' = √g'r'

where,

g' = 3g

V' = V

Therefore,

V = √3gr' ---------------- equation (2)

Comparing equation (1) and equation (2), we get:

√gr = √3gr'

gr = 3gr'

r = 3r'

r' = r/3

<u>Hence, the orbit radius would be divided by 3.</u>

5 0

3 years ago

Q C Ganymede is the largest of Jupiter's moons. Consider a rocket on the surface of Ganymede, at the point farthest from the pla

guapka [62]

The answer is $v=15.6 \mathrm{~km} / \mathrm{s}$ .

<h3>What is kinetic energy?</h3>

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

Determine the escape speed for the rocket from the planet-satellite system:

The potential energy of the rocket due to Ganymede when it is on the surface of the Ganymede is,

$U_1=-\frac{G M_{\mathrm{G}} m}{R_G}$

The potential energy of the rocket due to Jupiter

when it is on the surface of the Ganymede is,

$U_2=-\frac{G M_{\mathrm{J}} m}{R}$

Here, R is a separation between Jupiter and Ganymede.

To escape from the surface of Ganymede potential energy of the rocket due to Jupiter and Ganymede is equal to the kinetic energy of the rocket.

$\begin{aligned}&\frac{1}{2} m v^2+U_1+U_2=0 \\&\frac{1}{2} m v^2=-U_1-U_2 \\&\frac{1}{2} m v^2=\frac{G M_{\mathrm{G}} m}{R_G}+\frac{G M_{\mathrm{J}} m}{R} \\&\frac{1}{2} v^2=\frac{G M_{\mathrm{G}}}{R_G}+\frac{G M_{\mathrm{J}}}{R}\end{aligned}$

$v^2=\frac{2 G M_{\mathrm{G}}}{R_G}+\frac{2 G M_{\mathrm{T}}}{R}$

$v=\sqrt{2 G\left(\frac{M_{\mathrm{G}}}{R_G}+\frac{M}{R}\right)}$

$v=\sqrt{2\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right)\left(\frac{1.495 \times 10^{23} \mathrm{~kg}}{2.64 \times 10^6 \mathrm{~m}}+\frac{1.90 \times 10^{27} \mathrm{~kg}}{1.071 \times 10^9 \mathrm{~m}}\right)}$

$v=15.6 \times 10^3 \mathrm{~m} / \mathrm{s}$

$v=15.6 \mathrm{~km} / \mathrm{s}$

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

3 0

2 years ago