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2 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period

Physics

1 answer:

german2 years ago

3 0

2 minutes is 120 seconds, so if you were finding vibrations per minute, it would be 60 times a minute.

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The number of sound waves per unit time is called what

Andreyy89

Answer:

Frequency

Explanation:

Hoped this helped

3 0

3 years ago

A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc

hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

$E V = log_2(\dfrac{N^2}{t})$

N is the diameter of the aperture and t is the time of exposure

now,

$log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})$

$\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}$

inserting all the values

$\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}$

N₂² = 4

N₂ = 2 mm

hence , the correct answer is option E

4 0

3 years ago

The potential difference between the plates of an air-filled parallel-plate capacitor with a plate separation of 4 cm is 51 V. W

Salsk061 [2.6K]

Answer:

Explanation:

Electric field between plates of a parallel plate capacitor is uniform .

In a uniform electric field , relation between electric field and potential gradient is as follows

electric field = potential gradient [ E = - dV / dl ]

in the given case ,

dV = 51 V ,

dl = 4 cm

= 4 x 10⁻² m

E = 51 / 4 x 10⁻²

= 12.75 x 10² V / m

= 1275 V / m

6 0

3 years ago

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

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

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.

scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

Pressure at sea level = 1 atm = 101300 Pa

we know

P = ρ g h

$h = \dfrac{P}{\rho\ g}$

$h = \dfrac{101300}{1.3\times 9.8}$

h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

at x = 0 air density = 0

at x= h ρ_l = ρ_sl

assuming density is zero at x - distance

$\rho_x = \dfrac{\rho_{sl}}{h}\times x$

now, Pressure at depth x

$dP = \rho_x g dx$

$dP = \dfrac{\rho_{sl}}{h}\times x g dx$

integrating both side

$P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx$

$P =\dfrac{\rho_{sl}\times g h}{2}$

now,

$h=\dfrac{2P}{\rho_{sl}\times g}$

$h=\dfrac{2\times 101300}{1.3\times 9.8}$

h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0

3 years ago

A spring vibrates 120 times in 2 mins find its frequency and time period​

A spring vibrates 120 times in 2 mins find its frequency and time period