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Makovka662 [10]
3 years ago
15

If your weight is 588N on the earth, how far should you go from the centre of the earth so that your weight will be 300N? The ra

dius of the earth is 6400km and the value of g on the earth surface is 9.8m/s2. please explain....​
Physics
1 answer:
harkovskaia [24]3 years ago
4 0

Answer:

Explanation:

You need something that relates distance to what the gravitational pull is. You can set up a complex sort of proportion. What you need is a number that is comparable to 9.81 or you can just use the Gravitational Force formula with a 4 tier fraction.

Givens

x = the additional distance toward outer space above the radius of the earth.

G is the gravitational constant.

m1 = the person's mass (which does not change no matter where you are).

m2 = the earth's mass

F1 = 588 N

F2 = 300 N

Formula

\frac{F1}{F2} = \frac{588 N}{300N}=\frac{\frac{Gm1*m2/}{6400^2} }{\frac{G*m1*m2}{(6400 + x)^2} }

Solution

G*m1*m2 all cancel. So what you get looks like this.

\frac{588}{300} = \frac{(6400 + x)^2}{6400^2}

Cross Multiply

588 * 6400^2 = 300*(6400+x)^2  Now all you need do is solve for x.

x will be in km.

588*40960000 = 300 * (40960000 + 12800x + x^2)

1.2288*10^10 + 3840000x + 300x^2  = 2.408448*10^10

300x^2 + 3840000x + 1.2288*10^10 = 2.408448 * 10^10  

Subtract 2.409448 * 10^10 from both sides.

300x^2 + 3840000x - 1,179648 * 10^10

Now use the quadratic formula

I'm guessing I should have converted this to meters because I'm getting ridiculous numbers. They are already large enough as you can see. The method is correct, even if the numbers are not.  

 

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Answer:

The correct answer is from Tech B.

Explanation:

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Explanation:

Fluid gauge pressure is:

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where ρ is the fluid density and h is the depth of the fluid.

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Rounded to four significant figures, the gauge pressure is 16.09 MPa.

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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
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Answer:

V_1=8 V_2

Explanation:

Given that:

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We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

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An astronaut goes to Mars to do some experiments. Explain why her mass stays the same but her weight changes.
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A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?
belka [17]

Answer:

117.72kW

Explanation:

Given data

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height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

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Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

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