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3 years ago

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

Physics

1 answer:

GalinKa [24]3 years ago

7 0

Wave D has the same wavelength as wave A, but the amplitude is lower. The answer is Wave D.

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A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5\times 10^{-

joja [24]

Answer:

$5.25\cdot 10^{40} kg m^2/s$

Explanation:

The angular momentum of the pulsar is given by:

$L=m\omega r^2$

where

$m=2.8\cdot 10^{30} kg$ is the mass of the pulsar

$r = 10.0 km = 1\cdot 10^4 m$ is the radius

$\omega$ is the angular speed

Given the period of the pulsar, $T=33.5\cdot 10^{-3} s$ , the angular speed is given by

$\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s$

And so, the angular momentum is

$L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s$

8 0

3 years ago

Please I need help with these 2 questions. Thank you.

Lemur [1.5K]

First one is D and Second one is B

7 0

4 years ago

Light with a wavelength of 400 nm strikes the surface of cesium in a photocell, and the maximum kinetic energy of the electrons

Firdavs [7]

Answer:

The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

Explanation:

Given that,

Wavelength = 400 nm

Energy $E=1.54\times10^{-19}\ J$

We need to calculate the longest wavelength of light that is capable of ejecting electrons from that metal

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.54\times10^{-19}}$

$\lambda=1292\times10^{-9}\ m$

$\lambda=1292\ nm$

Hence, The longest wavelength of light that is capable of ejecting electrons from that metal is 1292 nm.

8 0

3 years ago

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

4 years ago

If the momentum of an object is doubled then kinetic energy is ...?

stiks02 [169]

Answer:

increased with the same rate as momentum

7 0

3 years ago