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Vesnalui [34]
3 years ago
8

Sally and Sam are in a spaceship that comes to within 17,000 km of the asteroid Ceres. Determine the force Sally experiences, in

N, due to the presence of the asteroid. The mass of the asteroid is 8.7 1020 kg and the mass of Sally is 80 kg. For calculation purposes, assume the two objects to be point masses.
Physics
1 answer:
MAXImum [283]3 years ago
8 0

Answer:

0.01606 Newtons

Explanation:

r = Distance between the asteroid and Sally = 17000000 m

m₁ = Mass of the asteroid = 8.7× 10²⁰ kg

m₂ = Mass of Sally = 80 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Newton's Universal law of gravity

F=G\frac{m_1m_2}{r^2}\\\Rightarrow F=6.67\times 10^{-11}\times \frac{8.7\times 10^{20}\times 80}{17000000^2}\\\Rightarrow F=0.01606\ N

The force Sally experiences is 0.01606 Newtons

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mars1129 [50]

Answer:

the answer is that the dough has the same mass before and after it was flattened

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2 years ago
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When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta
nikklg [1K]
When a river flows into an ocean, it slows down and deposits materials in its delta
8 0
3 years ago
A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi
inn [45]

Answer:

\Phi_{E} = E\pi r^2 \omega t

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

\Phi_{E} = \vec{E}\vec{A}

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

\Phi_{E} = EA\cos(\theta)

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

\theta = \omega t

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

\Phi_{E} = E\pi r^2 \omega t

3 0
3 years ago
The net electric charge of an amber rod which has been rubbed with fur is called negative Group of answer choices because amber
bonufazy [111]

Answer:

The right option is option E. None of the answer choices given are totally correct.

Explanation:

All insulators normally have an equal amount of positive and negative charges distributed on their surface.

The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.

The fur in turn becomes positive because it has more positive charges than negative on its surface.

So, the convention allows the now rubbed off amber rod to be called negative.

So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!

Hope this helps!!

3 0
3 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
inn [45]

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

6 0
3 years ago
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