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after an investigation kuri determines that her hypothesis was wrong what is the best thing for kuri to do next

aksik [14]

: You could make slight changes to the process or consider wheather the experiment was carried out correctly

7 0

3 years ago

What do we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect?Select

Sergio [31]

The work function is what we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect.

6 0

3 years ago

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The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

8 0

3 years ago

How are the mass and weight of an object related? Include a description with words and an equation.

nikdorinn [45]

The weight is the force experienced, whereas the mass represents the actual quantity of matter inside a body..
weigh on the surface of the earth is equal to mg
mass is m
and at the centre weight is 0 due to 0 acceleration that's 0 g
but mass is always constant and remains m, no matter where you are

6 0

3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago