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3 years ago

For a given wave IF frequency doubles the wavelength

Physics

1 answer:

motikmotik3 years ago

7 0

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

$v=f\lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

$\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}$

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A spring is 17 cm long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force th

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Answer:

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Explanation:

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2 years ago

Gravitational Force: Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravita

Alona [7]

Law of universal gravitation:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m = masses of the objects, r = distance between the objects

F is proportional to both M and m:

F ∝ M, F ∝ m

F is proportional to the inverse square of r:

F ∝ 1/r²

Calculate the scaling factor of F due to the change in M:

k₁ = 2M/M = 2

Calculate the scaling factor of F due to the change in m:

k₂ = 2m/m = 2

Calculate the scaling factor of F due to the change in r:

k₃ = 1/(4r/r)² = 1/16

Multiply the original force F by the scaling factors to obtain the new force:

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4 0

3 years ago

The critical angle for a liquid in air is 520, What is the liquid's index of refraction? 0.62 0.79 1.27 1.50

sammy [17]

Answer:

Liquid's index of refraction, n₁ = 1.27

Explanation:

It is given that,

The critical angle for a liquid in air is, $\theta_c=52^o$

We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.

Using Snell's law as :

$n_1sin\theta_c=n_2sin\theta_2$

Here, $\theta_2=90$

$sin\theta_c=\dfrac{n_2}{n_1}$

Where

n₂ = Refractive index of air = 1

n₁ = refractive index of liquid

So,

$n_1=\dfrac{n_2}{sin\theta_c}$

$n_1=\dfrac{1}{sin(52)}$

n₁ = 1.269

or n₁ = 1.27

Hence, the refractive index of liquid is 1.27

8 0

2 years ago

Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi

Snowcat [4.5K]

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ $V=20\ \Omega,\ V_1=30\ \Omega$

⇒ $G=1680\ \Omega,\ G_1=2930\ \Omega$

Concept to be used:

Conversion of galvanometer into voltmeter.

Let $G$ be the resistance of the galvanometer and $I_g$ the maximum deflection in the galvanometer.

To measure maximum voltage resistance $R$ is connected in series .

So,

⇒ $V=I_g(R+G)$

We have to find the value of $R$ we know that in series circuit current are same.

For $G=1680$ For $G_1=2930$

⇒ $I_g=\frac{V}{R+G}$ equation (i) ⇒ $I_g=\frac{V_1}{R+G_1}$ equation (ii)

Equating both the above equations:

⇒ $\frac{V}{R+G} = \frac{V_1}{R+G_1}$

⇒ $V(R+ G_1) = V_1 (R+G)$

⇒ $VR+VG_1 = V_1R+V_1G$

⇒ $VR-V_1R = V_1G-VG_1$

⇒ $R(V-V_1) = V_1G-VG_1$

⇒ $R =\frac{V_1G-VG_1}{(V-V_1)}$

⇒ Plugging the values.

⇒ $R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}$

⇒ $R =\frac{(50400 - 58600)}{(-10)}$

⇒ $R=\frac{-8200}{-10}$

⇒ $R=820\ \Omega$

The coil resistance of the circuit is 820 Ω .

4 0

2 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$