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IRISSAK [1]
3 years ago
5

The diagrams show two planets of different masses with identical orbiting satellites. Select all the conditions that would incre

ase the gravitational force between each pair.
A. move the satellites closer to the planet
B. move the satellites farther from the planet
C. Add mass to the satellites
D.Remove mass from the satellites
Physics
1 answer:
SIZIF [17.4K]3 years ago
8 0

Answer:

C I think but I'm not sure

Explanation:

Because by adding mass it would add more gravitational pull

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Which statement best describes the common features of sound waves and light waves
Viktor [21]
Sound waves or bounces off the wall and light waves are waves of light
7 0
3 years ago
Which of the statements best describes density?
Vesnalui [34]

Answer:

The answer to your question is: letter D.

Explanation:

a.The mass that a mole of substance has, measured in grams per mole.  Density is not measure in moles, so this is not the correct answer.

b.The amount of substance dissolved in a liquid, measured in moles per liter.  The substance dissolved in a liquid must be measure in grams not in moles, so this answer is incorrect.

c.The mass of substance dissolved in a liquid, measured in grams per milliliter.  I think that this definition is correct but is incomple, so this answer is wrong.

d.The ratio of a substance's mass to its volume, measured in grams per milliliter and also equivalent to grams per cubic centimeter. This is the right description to density, so this is the correct answer.

8 0
3 years ago
An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
A stone on ground is zero energy​
NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

<em>Hope</em><em> </em><em>this</em><em> helps</em><em> </em><em>!</em>

7 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
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