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2 years ago

5

the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its

angular acceleration in revolutions per minute squared

1 answer:

GalinKa [24]2 years ago

8 0

Answer:

please find attached pdf

Explanation:

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How many simple machines are in a wine opener? The kind with the bottle opener at the head and two "arms."

Blababa [14]

It uses the corkscrew to anchor it to the cork and a lever to pull the cork out

Hope this helps buddy:D

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2 years ago

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What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is doubl

vlabodo [156]

Answer:the force will remain same

Explanation:

because force is equal to the ratio of magnitude and distance

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2 years ago

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

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

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

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

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2 years ago

An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object

EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

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3 years ago

PLz I Need Help PLZZZ ASAP

AURORKA [14]

#1

As we are increasing the frequency in the simulation the wavelength is decreasing

So if speed remains constant then wavelength and frequency depends inversely on each other

If we are in boat and and moving over very small wavelengths then these small wavelength will be encountered continuously by the boat in short interval of times

#2

As we are changing the amplitude in the simulation there is no change in the speed frequency and wavelength.

So amplitude is independent of all these parameter

Amplitude of wave will decide the energy of wave

So light of greater intensity is the light of larger amplitude

#3

In our daily life we deal with two waves

1 sound waves

2 light waves

3 0

2 years ago