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2 years ago

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What region of the spectrum best corresponds to light with a wavelength equal to:_____ a. The diameter of a hydrogen atomb. The

size of a virus.c. Your height?

1 answer:

zavuch27 [327]2 years ago

3 0

We will make the comparison between each of the sizes against the known wavelengths.

In the case of the <em>hydrogen atom</em>, we know that this is equivalent to $10^{-10}$ m on average, which corresponds to the wavelength corresponding to X-rays.

In the case of the <em>Virus</em> we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.

In the case of <em>height</em>, it fluctuates in a person around $10 ^ 0$ to $10 ^ 1$ m, which falls to the wavelength of a radio wave.

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A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t

nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$ , Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V

4 0

3 years ago

Read 2 more answers

Two tuning forks, 254 Hz. and 260 Hz., are struck simultaneously. How many beats will be heard?

Lisa [10]

The longer you continue to listen, the more beats will be heard.

They'll occur at the rate of (260Hz - 254Hz) = 6 Hz .

7 0

3 years ago

The air in a kitchen has a mass of 60.0kg and a specific heat of 1505J/(kg°C).

BARSIC [14]

Your answer is 632,100J which is Choice D

8 0

3 years ago

Read 2 more answers

The distance between two planets is 1600 km. How much time would the light

Snowcat [4.5K]

Answer:

5.33*10^-3 seconds

Explanation:

c = d/t

c = speed of light constant (3.0*10^5 km/s)

d = distance (1600 km)

t = ?

3.0*10^5 = 1600/t

t = 1600/3.0*10^5

t = 5.33*10^-3 seconds

I hope this helped! :)

6 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago