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3 years ago

What do simple machines increase? Force Density Work Mechanical advantage

1 answer:

4 0

The wheel and axle increases your force. You exert your input force over a long distance and the output force is increased over a shorter distance. (Because the wheel is larger than the axle, the axle rotates and exerts a large output force.) A simple machine with a grooved wheel with a rope or cable wrapped around it.

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An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f

IgorC [24]

Resistance = (voltage) / (current)

Resistance = (6.0 v) / (2.0 A)

Resistance = 3.0 ohms

7 0

3 years ago

Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in t

noname [10]

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

<em>The phenomenon of Doppler effects is generally applicable to both sound and light. </em>

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

<em>Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc. </em>

3 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

How do you know the speed of an electromagnetic wave in a vacuum?

ycow [4]

Electromagnetic waves need no matter to travel - they can travel through empty space (a vacuum). In a vacuum, all electromagnetic waves travel at approximately 3 x 108 m/s - which is the fastest speed possible. ...
Light traveling value through an optical Fibre is, 2 x 108 m/s. Hope that helps.

5 0

3 years ago

You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m

ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

$\sf{h = \frac{1}{2} \cdot g \cdot t^2}$

$\sf{\frac{2 \cdot h}{g} = t^2}$

$\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}$

With the following condition :

t = interval of the time (s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

$\boxed{\sf{\bold{t = \frac{s}{v}}}}$