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Minchanka [31]
3 years ago
6

What is the magnification of an object that is 4.15 m in front of a camera that has an image position of 5.0 cm?

Physics
1 answer:
Stolb23 [73]3 years ago
4 0
-0.012

Done !!!!!!!!
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The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
3 years ago
What can you say about the magnitudes of the forces that the balloons exert on each other?
maxonik [38]

Answer:

F_G=G. \frac{m_1.m_2}{R^2} gravitational force

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2} electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

F_G=G. \frac{m_1.m_2}{R^2}

where:

G= gravitational constant  =6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}

m_1\ \&\ m_2 are the masses of individual balloons

R= the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}

where:

\rm \epsilon_0= permittivity\ of\ free\ space

q_1\ \&\ q_2 are the charges on the individual balloons

R = radial distance between the charges.

3 0
3 years ago
A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?​
ryzh [129]

Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

s = ut + 1/2 at²

since the acceleration of sound is 0:

s = ut + 1/2 (0)t²

s = ut    <em>(here, u is the speed of sound , s is the distance travelled and t is the time taken)</em>

Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0
3 years ago
What is the weight of an astronaut with a mass of 75 kg on the moon? The gravitational field strength on the moon is 1.6 Nkg-1.
bearhunter [10]

okay this is kinda easy

<u>What is the gravitational field strength on the moon?</u>

The Moon has a gravitational field strength of 1.6 N/kg.

7 0
3 years ago
Two point charges, initially 2 cm apart, are moved to a distance of 10 cm apart. By what factor does the resulting electric forc
choli [55]

Answer: The correct answer is C.

2/10 is simplified to 1/5, that is the factor that changes.

5 0
3 years ago
Read 2 more answers
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