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marissa [1.9K]
2 years ago
12

How is air resistance similar to gravity? give me two ways.

Physics
1 answer:
OLga [1]2 years ago
4 0

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

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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
Mr. MacDougall got his vehicle stuck in the snow. Being the nice student that you are, you stop to help Mr. MacDougall out of th
zhannawk [14.2K]

Answer:

Explanation:

work done=force*displacement

=350N*15m

=5250 joule

4 0
2 years ago
If you do 1500 J of work hoisting a 20 kg bale of hay , to what height did you lift it
strojnjashka [21]

Explanation:

W = PE

W = mgh

1500 J = (20 kg) (9.8 m/s²) h

h = 7.65 m

Round as needed.

4 0
3 years ago
If the trend changed toward traditional (pre-World War II) families, how would that affect women’s rights?
LUCKY_DIMON [66]

If the trend changed toward traditional (pre-World War II) families, the women’s rights are employment in manufacturing sector.

<h3>What is World war?</h3>

The war between two countries to take over each other's kingdom using weapons to kill each other.

Before the world war, the army needs armor, weapons, guns and tanks. Their manufacturing is only possible with many workers to work for long hours. If the men are not enough, then women are given opportunities to work with them.

Thus, the women’s rights are employment in manufacturing sector when trend changed traditional families.

Learn more about world war.

brainly.com/question/1449762

#SPJ1

7 0
2 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
3 years ago
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