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3 years ago

A 150 cm long string vibrates with 3 loops and its frequency is 80 Hz. What will be the wavelength and velocity of the waves?

Physics

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

since each loop is ewuvivalent to one half wave lenght . the length of the string is equal to two halves of a wavelength . put in the form of an equation in the same reasoning also

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The resolving power of a microscope is proportional to the wavelength used. A resolution of 1.0 10-11 m (0.010 nm) would be requ

Anon25 [30]

Answer:

K = 13448.64eV

Explanation:

(a) In order to calculate the kinetic energy of the electrons, to "see" the atom, you take into account that the wavelength of the electrons must be of the order of the resolution required (0.010nm).

Then, you first calculate, by using the Broglies' relation, the momentum of the electron associated to a wavelength of 0.010nm:

$p=\frac{h}{\lambda}$ (1)

p: momentum of the electron

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 0.010nm

You replace the values of the parameters in the equation (1):

$p=\frac{6.262*10^{-34}Js}{0.010*10^{-9}m}=6.262*10^{-23}kg\frac{m}{s}$

With this values of the momentum of the electron you can calculate the kinetic energy of the electron by using the following formula:

$K=\frac{p^2}{2m}$ (2)

m: mass of the electron = 9.1*10^-31 kg

$K=\frac{(6.262*10^{-23}kgm/s)^2}{2(9.1*10^{-31}kg)}=2.15*10^{-15 }J$

In electron volts you obtain:

$2.15*10^{-15}J*\frac{6.242*10^{18}eV}{1J}=13448.64eV$

The kinetic energy required for the electrons must be, at least, of 13448.64 eV

5 0

3 years ago

A device used to increase or decrease the emf in the second of two unconnected coils is a

grigory [225]

The answer is transformer. They utilize electromagnetic induction to generate current. This is only possible in alternating current due to the differential increase and decrease of electrical current that induces changes in magnetic flux in the coil. This varies the magnetic flux of the primary coil that generates current in the secondary coil.

4 0

3 years ago

In a 35 mm single lens reflex camera (SLR) the distance from the lens to the film is varied in order to focus on objects at vary

Evgesh-ka [11]

Answer:

range of movement is 1.49 mm

Explanation:

given data

focal length = 45 mm

distance = 1.4 m

distance from the lens = 35 mm

distance from infinity down = 1.4 m =

to find out

range of movement

solution

we will apply here lens equation that is

1/f = 1/p + 1/q

here f = 45 and p = infinity to 1400 mm

we find here image distance that is q

1/45 = 0 + 1/q ......1

q = 45 mm

and

1/45 = 1/1400 + 1/q ......2

q = 46.49

so range of movement

that is 46.49 - 45

range of movement is 1.49 mm

8 0

3 years ago

A car slows down uniformly from a speed of 22 m/s to rest in 4.0 seconds. How far did it travel in that time

krek1111 [17]

Answer:

$\boxed{\sf Distance \ travelled = 44 \ m}$

Given:

Initial speed (u) = 22 m/s

Final speed (v) = 0 m/s (Rest)

Time taken (t) = 4 seconds

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = (\frac{v + u}{2} )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( \frac{0 + 22}{2} ) \times 4 \\ \\ \sf \implies s = \frac{22}{2} \times 4 \\ \\ \sf \implies s = 11 \times 4 \\ \\ \sf \implies s = 44 \: m$

$\therefore$

Distance travelled by car (s) = 44 m

4 0

4 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

The range is 35.35 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

3 years ago