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artcher [175]
3 years ago
6

Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb

it, assuming a circular orbit?
Physics
1 answer:
Maslowich3 years ago
5 0

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

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Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab
forsale [732]

Answer:

232.9m³ (Option b. is the closest answer)

Explanation:

Given:

Air pressure in the lab before the storm, P₁ = 1.1atm

Air volume in the lab before the storm, V₁ = 180m³

Air pressure in the lab during the storm P₂ = 0.85atm

Air volume in the lab before the storm, V₂ = ?

Applying Boyle's law:    P₁V₁ = P₂V₂    (at constant temperature)

                          V_{2} = \frac{P_{1}V_{1}}{P_{2}}

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                          V_{2} = \frac{198}{0.85}

                           V₂  = 232.9m³

The air volume in the laboratory that would expand in order to make up for the large pressure difference outside is 232.9m³

7 0
3 years ago
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.2589 N when separated by
Harlamova29_29 [7]

Answer:

q_1 = 7.19 \times 10^{-6} C

q_2 = -1.0 \times 10^{-6} C

Explanation:

Let the initial charge on the two spheres are

q_1, -q_2

now we know that the force between them is given as

F = \frac{kq_1q_2}{r^2}

0.2589 = \frac{kq_1q_2}{0.5^2}

q_1q_2 = 7.19 \times 10^{-12}

now when two spheres are connected then final charge on them is given as

q = \frac{q_1 - q_2}{2}

now the force between them is given as

0.3456 = \frac{k(q_1 - q_2)^2}{4(0.5)^2}

now we have

q_1 - q_2 = 6.19 \times 10^{-6}

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q_1 = 7.19 \times 10^{-6} C

q_2 = -1.0 \times 10^{-6} C

3 0
2 years ago
The unit for measuring the rate at which light energy is radiated from a source is the
Marizza181 [45]
The unit for measuring the rate at which light is radiated from a source is B, Lumen. Lumenous flux is the time rate of the flow of light, or the visible energy produced from a certain light source.  It is the quantitative measure of brilliance of a light source. The unit we use for this is lumen. 
8 0
2 years ago
What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
2 years ago
A tank circuit consists of an inductor and a capacitor. Give a simple explanation for why the magnetic field in the induc- tor i
ipn [44]

Answer:

If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?

Charge will oscillate in the tank's capacitor and inductor.

Explanation:

4 0
3 years ago
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