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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....

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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Artist 52 [7]

Answer:

The distance of the first bright fringe is given as $Y_C = 1.22 *10^{-3}m$

The distance of the second dark fringe from the central bright fringe is given as $Y_D = 0.00192 \ m$

Explanation:

From the question we are told that

The slit separation distance is $d = 1.15 mm = \frac{1.15}{1000} =0.00115 m$

The distance of the slit from the screen is $D = 3.23 m$

The wavelength is $\lambda = 633 nm$

For constructive interference to occur the distance between the two slit is mathematically represented as

$Y_C =\frac{m \lambda D}{d}$

Where m is the order of the fringe which has a value of 1 for first bright fringe

Substituting values

$Y_C = \frac{1 * 633 *0^{-9} * 3.23}{0.00115}$

$Y_C = 1.22 *10^{-3}m$

For destructive interference to occur the distance between the two slit is mathematically represented as

$Y_D = [n + \frac{1}{2} ] \frac{\lambda D}{d}$

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so the formula to get the dark fringe is $n = \frac{1}{2} * 1$

$n=1$

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$Y_D = [ 1 + \frac{1}{2} ] * \frac{633 *10^{-9} * 3.23 }{0.00115}$

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3 years ago

A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera

AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel $r=0.3\ m$

Initial angular velocity $\omega _0=0$

It rotates 3 revolution in 5 s therefore

$\omega =2\pi 3=\6\pi =18.85\ rad/s$

using $\omega =\omega _0+\alpha t$

where $\alpha =angular\ acceleration$

$\omega =Final\ angular\ velocity$

$t=time$

$\alpha =\frac{18.85}{5}=3.77 rad/s^2$

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

$\omega at t=1$

$\omega =0+3.77\times 1=3.77 rad/s$

$a_c=\omega ^2\cdot r$

$a_c=(3.77)^2\cdot 0.3=4.26 m/s^2$

Tangential acceleration $a_t=\alpha \times r$

$a_t=3.77\times 0.3=1.13 m/s^2$

$a_{net}=\sqrt{a_t^2+a_c^2}$

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3 years ago

a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro

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Answer:

<h2>73 kg</h2>

Explanation:

The mass of the object can be found by using the formula

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<h3>73 kg</h3>

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3 years ago