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2 years ago

9

What variable is represented on the y-axis?

1 answer:

gladu [14]2 years ago

8 0

Distance/ Time which means Distance is on horizontal and time is on vertical

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When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point

vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

$v = Velocity \rightarrow 20m/s$

$\lambda = Wavelength \rightarrow 35*10^{-2}m$

Therefore the frequency would be given as

$f = \frac{20}{35*10^{-2}}$

$f = 57.14Hz$

The frequency is directly proportional to the angular velocity therefore

$\omega = 2\pi f$

$\omega = 2\pi *57.14$

$\omega = 359.03rad/s$

Now the maximum speed from the simple harmonic movement is given by

$V_{max} = A\omega$

Where

A = Amplitude

Then replacing,

$V_{max} = (1*10^{-2})(359.03)$

$V_{max} = 3.59m/s$

Therefore the maximum speed of a point on the string is 3.59m/s

8 0

3 years ago

Which term refers to the bending of light rays when they encounter the edge of a barrier?

MatroZZZ [7]

Answer:

Diffraction

Explanation:

The bending of light around obstacle called diffraction.

4 0

3 years ago

Read 2 more answers

Help not to sure with this one need help plz asap

DiKsa [7]

A is the answer for the problem

4 0

2 years ago

Read 2 more answers

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago

We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st

anastassius [24]

Solution :

a). Using Gauss's law :

$E=\frac{Q}{4 \pi \epsilon_0r^2}$ , $$b$ .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$ .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

$=\int^a_b E \ dx$

$=\frac{Q}{4 \pi \epsilon_0} \int^a_b \frac{1}{x^2} \ dx$

$=\frac{Q(a-b)}{4 \pi \epsilon_0 a b}$ ....................(3)

Therefore, $U=\frac{1}{2}Q \Delta V$

$=\frac{1}{2}(4 \pi \epsilon_0 b^2 E_0)\left(\frac{Q(a-b)}{4\pi \epsilon_0 a b}\right)$

$=\frac{4 \pi \epsilon_0}{2a} \ E^2_0 b^3(a-b)$ .............(4)

Now differentiating the equation (4) w.r.t. 'b', we get

$b=\frac{3}{4}a$

Thus the radius for the inner cylinder conductor is $b=\frac{3}{4}a$

b). For the energy storage, substitute the radius in (4), we get

$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

This is the amount of energy stored in the conductor.

8 0

2 years ago