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Alona [7]
3 years ago
9

Philosophy: The Big Picture Unit 8

Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
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a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi
e-lub [12.9K]

Answer:

en español por

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4 0
2 years ago
If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?
stiks02 [169]

Answer:

1 W = 1 J / sec       Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min    energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

4 0
2 years ago
The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length
Zarrin [17]

The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

y_r=1.22\frac{\lambda f}{d}

Where,

y_r = Range of the radius

\lambda = wavelength

f= focal length

Our values are given by,

State 1:

d=2.00mm = 2*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 750nm = 750*10^{-9}m

State 2:

d=8.00mm = 8*10^{-3}m

f= 25mm = 25*10^{-3}m

\lambda = 390nm = 390*10^{-9}

Replacing in the first equation we have:

y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}

y_{r1}= 11.4\mu m

And also for,

y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}

y_{r2} = 1.49\mu m

Therefor, the airy disk radius ranges from 1.49\mu m to 11.4\mu m

7 0
3 years ago
A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca
UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

6 0
3 years ago
1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
3 years ago
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