Ask question

Ask question

All categories

2 years ago

13

Frequency definition class 9 sub science

2 answers:

tia_tia [17]2 years ago

7 0

Explanation:

1.

the rate at which something occurs over a particular period of time or in a given sample.

"an increase in the frequency of accidents due to increased overtime"

2.

the rate per second of a vibration constituting a wave, either in a material (as in sound waves), or in an electromagnetic field (as in radio waves and light).

"different thicknesses of glass will absorb different frequencies of sound"

Vedmedyk [2.9K]2 years ago

3 0

Answer:

Frequency is the number of occurrences of a repeating event per unit of time.

You might be interested in

Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da

FrozenT [24]

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light $\lambda =550nm=550\times 10^{-9}m$

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So $x=3\times 10^{-3}m$

We have to find the width of the slit

For the first order wavelength is equal to $\lambda =\frac{x}{D}\times a$ , here a width of slit

So $a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm$

So width of the slit will be equal to 1.47 mm

5 0

3 years ago

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

2 years ago

Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency

Scilla [17]

Answer:

A and B

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength$

c is the speed of light having value $3\times 10^8\ m/s$

Thus, the product of the wavelength and the frequency is constant and equal to $3\times 10^8\ m/s$

<u>Option A is correct.</u>

Given, Frequency = $1\times 10^{18}\ Hz$

Thus, Wavelength is:

$Wavelength=\frac{c}{Frequency}$

$Wavelength=\frac{3\times 10^8}{1\times 10^{18}}\ m$

$Wavelength=3\times 10^{-10}\ m$

Also, 1 m = $3\times 10^{-10}$ Å

So,

<u>Wavelength = 3.0 Å</u>

<u>Option B is correct.</u>

As stated above, the speed of electromagnetic radiation is constant. Hence, each radiation of the spectrum travels with same speed.

<u>Option C is incorrect.</u>

3 0

3 years ago

Read 2 more answers

What type of magnetism is present in a magnet that you use to place a photograph on the refrigerator?

Vika [28.1K]

Answer:

C) Contact

Explanation:

The magnet requires almost direct <u>contact</u> with the fridge to start its magnetic properties.

7 0

3 years ago

Read 2 more answers

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

2 years ago