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marshall27 [118]
3 years ago
13

Will a roller coaster with a higher starting point be a faster ride? Why or why not

Physics
1 answer:
Brilliant_brown [7]3 years ago
8 0

Answer:

yez

Explanation:

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A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875
mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0
3 years ago
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5. The Weeks family took a trip to the Dallas 200 for
zheka24 [161]

Answer: idk

Explanation: hahahaha

4 0
3 years ago
On which date is the gravitational force between Earth and the moon the greatest?
Genrish500 [490]

The gravitational forces between the Earth and Moon are greatest when the two bodies are closest together. That happens every 27.32 days, when the Moon is at the perigee of its orbit.

Even if this happened at the same time in every orbit, the date would change, because there are not 27.32 days in a month.

But it doesn't happen at the same time in every orbit ... the Moon's perigee precesses around its orbit, on account of the gravitational forces toward the Earth, the Sun, Venus, Mars, and the other planets.

3 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
What kind of elements tend to have higher ionization energies.
algol13
Non metals tend to have higher ionization energies
5 0
2 years ago
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