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klio [65]
3 years ago
8

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

Physics
1 answer:
siniylev [52]3 years ago
7 0

Answer:

The height of the  object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of Gmm_{e}

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

200=\dfrac{Gmm_{e}}{(3958.756)^2}

200\times(3958.756)^2=Gmm_{e}

Gmm_{e}=3.134\times10^{9}

We need to calculate the height of the  object

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

125=\dfrac{200\times(3958.756)^2}{r^2}

r^2=\dfrac{200\times(3958.756)^2}{125}

r^2=25074798.5

r=\sqrt{25074798.5}

r=5007.4\ miles

Hence. The height of the  object is 5007.4 miles.

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Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

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● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

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■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

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■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

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