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gogolik [260]
3 years ago
13

Why do the birds and butterflies keep flying into my window? (4 butterflies, 2 birds). I think there after my blood.

Physics
1 answer:
Zina [86]3 years ago
6 0

Answer:

Is that an actual question for school?

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force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it
Yanka [14]

Answer:

Given force=10lb

L1=4in converting to feet

But 0.08333ft= 1 inch

Then 4 inch is 0.3332

6inch is 0.49998

But hookes law states

F=Kx where F is force,K is the force constant ,X

K=F/X=10/0.3333=30N/m

Integrating this

Integral of 30x with limit 0.333 to 0.5

F=30x^2/2=15x^2substing the limit

F=(15(0.5^2-0.33^2)=2.08ft-lb

Explanation:

3 0
3 years ago
What goes through evaporation
Olegator [25]
Water goes through evaporation.
6 0
3 years ago
Read 2 more answers
Unit 12 Exam
lapo4ka [179]

Answer:

You increase the acceleration of the car.

6 0
2 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

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

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
2 years ago
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