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2 years ago

10

By what factor would your weight be multiplied if earth's diameter were 1/ 4 as big and earth's mass remained unchanged?.

1 answer:

GenaCL600 [577]2 years ago

3 0

Answer:

16

Step-by-step explanation:

Your weight is (roughly) proportional to the inverse of the square of the distance you are from the center of the earth. If that distance is reduced to 1/4 its previous value, then your weight will be multiplied by the factor ...

1/(1/4)² = 16

You would weigh 16 times as much.

_____

<em>Additional comment</em>

The distribution of mass within the earth also affects your weight.

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What is an advantage that experimental probability has over theoretical probability?

DIA [1.3K]

C. It is not always possible to calculate theoretical probability in complex situations, but experimental probability can be calculated any time experiments can be run.

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3 years ago

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The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

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3 years ago

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The length of a rectangular piece of land is 92 yards more than three times its width. The perimeter is 760 yards. find its dime

wlad13 [49]

Answer: Dimensions are given as below:

Length = 236 yards

width = 144 yards

Step-by-step explanation:

Let the width be x

then length = 92 more than x = 92+x

formula for perimeter of rectangle = 2 [ length + width]

therefore

760 = 2 [ 92+x + x ]

760 = 2 ( 92+2x )

760 = 184 + 4x

4x = 760 - 184

4x = 576

x= $\frac{576}{4}$ =

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So width = 144 yards

Length = 144+92 = 236 yards

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Answer:

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Step-by-step explanation:

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Answer:

Pounds

Step-by-step explanation:

An airplane is so heavy

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