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3 years ago

A pendulum is in a spacecraft to measure

Physics

1 answer:

Tcecarenko [31]3 years ago

4 0

Answer:

For small angles (✓ < 30 degrees), the period of a simple pendulum4can be approximated by: 3A radian is an angle measure based upon the circumference of a circle C =2⇡r where r is the radius of the circle. A complete circle (360 ) is said to have 2⇡ radians. Therefore, a 1/4 circle (90 )is⇡/2 radians.

Explanation:

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How many atoms of oxygen are there in one molecule of trinitrotoluene, if the chemical formula is C6H3(NO2)3?

kvasek [131]

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2 x 3 = 6

Thus there are 6 atoms of oxygen in this molecule.

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6 0

3 years ago

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The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre

Dmitrij [34]

M represent mass in kg
v represents speed in m/s
r represents radius in m

Now, just substitute these into the formula:
 $F = \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }$

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4 years ago

In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?

gizmo_the_mogwai [7]

It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left

5 0

3 years ago

Read 2 more answers

What is the kind of energy that comes from movement?

Sergio [31]

Answer:

motion energy