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3 years ago

Please help i really need it. this is my physic exam

Physics

2 answers:

erastova [34]3 years ago

7 0

Correct answer is C.

Harlamova29_29 [7]3 years ago

3 0

Answer:

bout=14.4V , b)R2=80

am i right

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Una ave vuela a una velocidad constante de 15m/s en una trayectoria rectilínea. Si dura una hora volando ¿cuanta distancia habrá

earnstyle [38]

Answer:

54,000

Explanation:

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3 years ago

Mr. Patel is looking for his cat in the garden at night. It is completely dark outside. Mr Patel shines his torch at the tree. W

UNO [17]

Answer:

the tree's light reflects into his eyes.

I think this is the ans

Explanation:

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2 years ago

The sound of a tuba is very low. Why?

vaieri [72.5K]

Answer:

When you blow into a tuba the air vibrates very slowly.

Explanation:

Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .

Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.

8 0

3 years ago

Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

marysya [2.9K]

$Volume of balloon = \frac{4}{3} \pi r^{3}$

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V = 68+ 0.178*V

0.934 V = 68

V = 72.81 $m^3$

\frac{4}{3} \pi r^{3}[/tex]= 72.81

r = 2.59 m

So radius of helium balloon = 2.59 m

4 0

3 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

3 years ago