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DaniilM [7]
3 years ago
7

Which point on the standing wave is a node?

Physics
2 answers:
Digiron [165]3 years ago
8 0

Answer: it’s c

Explanation: ap3x

Evgesh-ka [11]3 years ago
5 0

Answer:

B is right on A.P.E.X

Explanation:

search up node and it'll show you

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Suppose that a ball is released from the window of a train that is moving with constant velocity.  The path of the ball, as obse
barxatty [35]

True, the path of the ball, as observed from the train window, will be a horizontal straight line.

An object projected from a certain height has a parabolic path when observed from a fixed point.

However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.

When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.

Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.

Learn more about path of motion of objects here: brainly.com/question/82610

7 0
2 years ago
Accelerating charges radiate electromagnetic waves. Calculate the wavelength of radiation produced by a proton of mass mp moving
Iteru [2.4K]

Explanation:

Let m_p is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

\dfrac{mv^2}{r}=qvB

r is the radius of path,

r=\dfrac{mv}{qB}

Time period is given by :

T=\dfrac{2\pi r}{v}

T=\dfrac{2\pi m_p}{qB}

Frequency of proton is given by :

f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}

The wavelength of radiation is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{2\pi m_pc}{qB}

So, the wavelength of radiation produced by a proton is \dfrac{2\pi m_pc}{qB}. Hence, this is the required solution.

3 0
3 years ago
1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

8 0
3 years ago
I need help with 1-7 ASAP
elena55 [62]

Refraction. ... Diffraction. ... EM spectrum. ... Intensity. ... Transverse wave. ... Frequency. ... Compression wave.

5 0
4 years ago
A planet exerts a gravitational force of magnitude 4e22 N on a star. If the planet were 3 times closer to the star (that is, if
Alex_Xolod [135]

Answer:

3.6\times10^{23} N

Explanation:

F=\frac{GmM}{r^2}=4\times10^{22} N

F'=\frac{GmM}{(r/3)^2}=9\frac{GmM}{r^2}=9\times4\times10^{22}=3.6\times10^{23} N

7 0
2 years ago
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