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3 years ago

10

Plz help me ASAP How does sound travel from its source to your ear??? 1. By changes in air pressure 2. By vibrations in wires or

strings 3. By electromagnetic wave 4. By infrared waves

1 answer:

Sholpan [36]3 years ago

8 0

Answer:

2. By vibrations in wires or strings.

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When light hits a mirror it bounces off of it. This is why you can see yourself in a mirror. This property of light is A) absorp

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Based on the path of a light ray through air, water, and glass, which medium has the greatest index of refraction?

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Glass is the answer for this question

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A 25 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction

Lunna [17]

Up until the moment the box starts to slip, the static friction is maximized with magnitude <em>f</em>, so that by Newton's second law,

• the net force acting on the box parallel to the ramp is

∑ <em>F</em> = <em>mg</em> sin(<em>α</em>) - <em>f</em> = 0

where <em>mg</em> sin(<em>α</em>) is the magnitude of the parallel component of the box's weight; and

• the net force acting perpendicular to the ramp is

∑ <em>F</em> = <em>n</em> - <em>mg</em> cos(<em>α</em>) = 0

where <em>n</em> is the magnitude of the normal force and <em>mg</em> cos(<em>α</em>) is the magnitude of the perpendicular component of weight.

From the second equation we have

<em>n</em> = <em>mg</em> cos(<em>α</em>)

and <em>f</em> = <em>µn</em> = <em>µmg</em> cos(<em>α</em>), where <em>µ</em> is the coefficient of static friction. Substituting these into the first equation gives us

<em>mg</em> sin(<em>α</em>) = <em>µmg</em> cos(<em>α</em>) ==> <em>µ</em> = tan(<em>α</em>) ==> <em>α</em> = arctan(0.35) ≈ 19.3°

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2 years ago

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

3 years ago

You record the spectrum of a star and find that a calcium absorption line has an observed wavelength of 394.0 nm. This calcium a

saveliy_v [14]

Answer:

radial velocity = 0.533 × $10^{6}$ m/s

Explanation:

given data

observed wavelength = 394.0 nm

rest wavelength = 393.3 nm

solution

we get here radial velocity by the Doppler effect of light that is

$\frac{\triangle \lambda }{\lambda} = \frac{v}{c}$ ......................1

v = $\frac{c \times \triangle \lambda }{\lambda}$

put here value

v = \frac{c \times (\lambda ' - \lambda }{\lambda}

v = \frac{3 \times 10^8 \times (394 - 393.3}{393.3}

v = 533943.55

v = 0.533 × $10^{6}$ m/s

8 0

3 years ago