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Gre4nikov [31]
3 years ago
7

A diver 50 m deep in 10∘C fresh water exhales a 1.0-cm-diameter bubble. What is the bubble's diameter just as it reaches the sur

face of the lake, where the water temperature is 20∘C?
Physics
1 answer:
aliya0001 [1]3 years ago
4 0

Answer:

18.2mm

Explanation:

D = 50m

T1 = 10+273 = 283K

T2 = 20+273 = 293K

R1 = 5x10^-3

Absolute pressure at 50m

P1 = pA + pwateer x g x d

= 101000+ 1000x9.81x50

= 591500pa

New volume of bubble

= P1v1/T1 = p2v2/T2

= 125x10^-9 x 591500x293/101000*283

= 757.9x10^-9m³

R2 = 9.2x10^-3

D2 = 18.2mm

Or 1.82cm

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What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?
lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34<span> m2 kg / s).The momentum is: 2.2 x 10^(-25).</span>

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31<span> kilograms).</span>

v=2.2 x 10^(-25)/9.10938356 × 10-31<span> kilograms=0.24 x 10^6 m/s</span>

8 0
3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
4 years ago
A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy
Vsevolod [243]

Answer:

v = 4.2 \ m/s

Explanation:

Given data:

Mass of the paper clip, m = 1.5 \ g = 0.0015 \ kg

Kinetic energy, K = 0.013 \ \rm J

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

K = \frac{1}{2}mv^{2}

0.013 = 0.5 \times 0.0015 \times v^{2}

\Rightarrow \ v = 4.16 \ m/s

v = 4.2 \ m/s.  (rounding to nearest tenth)

3 0
3 years ago
Suppose you look at a spectrum of visible light by looking through a prism or diffraction grating. How can you decide whether it
elena55 [62]

Answer:An emission line spectrum consists of bright lines on a dark background, while an absorption line spectrum consists of dark lines on a rainbow background.

Explanation:

5 0
3 years ago
Read 2 more answers
A student estimated a mass to be 325 g, but the actual mass is 342 g. What is the percent error?
Dovator [93]

Answer:

Answer:

5.2307 %

Explanation:

(acutal mass- estimated mass) / ( estimated mass)

3 0
3 years ago
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