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3 years ago

When a wave enters a new medium, it

1 answer:

5 0

Velocity of a wave = wavelength * frequency

The velocity may change and the wavelength may change--- the frequency stays constant!

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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil

tatyana61 [14]

Answer:

The time taken by missile's clock is $4.6\times 10^{6} s$

Solution:

As per the question:

Speed of the missile, $v_{m = 6.5\times 10^{3}} m/s$

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

$T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}$

$1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}$ (1)

Using binomial theorem in the above eqn:

We know that:

$(1 + x)^{a} = 1 + ax$

Thus eqn (1) becomes:

$1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}$

$T = \frac{2c^{2}}{v_{m}^{2}}$

Now, putting appropriate values in the above eqn:

$T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}$

$T = 4.6\times 10^{6} s$

4 0

3 years ago

If object a has more mass than object b does object a contain more matter?

aev [14]

The answer should be a yes

8 0

3 years ago

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$