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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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2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in
aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

To learn more about Work Done visit:

https://brainly.in/question/48222628

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2 years ago
A pharmacist attempts to weigh 100 milligrams of codeine sulfate on a balance with a sensitivity requirement of 4 milligrams. Ca
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Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100)  x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

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3 years ago
Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all
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wow that is confusing


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Which of the following is equal to Velocity change per second​
Mademuasel [1]

Acceleration

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Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

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V is the initial velocity

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8 0
3 years ago
Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a
nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

Q = \int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0  \int\limits^r_0 {r^3} \, dr * d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi

Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \,  d\phi

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Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0
3 years ago
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