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pav-90 [236]
3 years ago
10

The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

Physics
1 answer:
n200080 [17]3 years ago
3 0
<h2>Answer:g=9.79ms^{-2},A object of mass m at the surface of earth experiences a force mg</h2>

Explanation:

Let M be the mass of earth.

Let R be the radius of earth.

Let G be the universal gravitational constant.

Given,

M=5.96\times 10^{24}Kg

R=6.37\times 10^{6}m

G=6.67259 \times 10^{-11}Nm^{2}Kg^{-2}

Let g be the acceleration due to gravity.

Then,g=\dfrac{GM}{R^{2}}

g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}

g=9.79ms^{-2}

A object of mass m at the surface of earth experiences a force mg

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6 0
3 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
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4 0
3 years ago
How and where is old oceanic crust destroyed?
sladkih [1.3K]

Answer:

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Explanation:

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8 0
3 years ago
Read 2 more answers
Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?
Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

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To get her typing speed per minute, we will use the formula

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6 0
3 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

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