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3 years ago

11

2. A wave on a rope has a wavelength of 2.0 m and a frequency of 2.0 Hz. What is the speed of the

1 answer:

TEA [102]3 years ago

7 0

Answer:

4 m/s or 4 meters per second.

Explanation:

In order to calculate the speed of wave, you multiply the wavelength in meters and the frequency of the Wave in Hertz. 2 times 2 equals 4. The wave speed is always in m/s considering that the wavelength is also in meters.

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Why is centrifugal force is a fake force?<br>

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Answer:

We say fictitious because the actual source of the centrifugal acceleration is somewhat indirect and the experience one has results from the unbalanced forces acting on the reference frame, not a force. Note, it is an acceleration not a force. For instance, imagine yourself on a swing.

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A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.

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3 years ago

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

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Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

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3 years ago

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

kotegsom [21]

Answer:

t = 16.94 s

Explanation:

t is the time passes before police catch the speeder

speed of speeder Vo = V = 23.3 m/s

T = t

Police Info

Vo = 0 m/s

a = 2.75 m/s^2

t = t

Now,

displacement of the police car = displacement of the speeder.

x_{police} = Vo *t + 1/2 at^2

since Vo = 0

x police = 1/2 at^2

x police = 1/2 (2.75)(t)^2

Now the displacement of speeder is

x_{speeder} = Vt

x_{speeder} = 23.3 t

x_{speeder} = x_{police}

23.3 t = 1/2 * 2.75 t^2

23.3 t = 1.375 t^2

t = 23.3\1.375

t = 16.94

t = 16.94 s

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