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fiasKO [112]
3 years ago
9

5. What is the amplitude and wavelength of the wave?

Physics
1 answer:
scZoUnD [109]3 years ago
5 0

Answer:

Amplitude = 5 cm

Wavelength = 30 m

Explanation:

Amplitude is the maximum displacement. According to the graph, the maximum height is the maximum displacement which is equal to 5cm.

Wavelength is the same distance between two successive Crest or trough.

From the graph the wavelength is 30 metres.

That is,

Amplitude = 5 cm

Wavelength = 30 m

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To the right

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Explanation:

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2 years ago
The speed of molecules is indirectly related to a measurement called _____.
seraphim [82]

The average speed of all the molecules in an object
or sample of a substance is related to its temperature ...
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An RV is traveling 60 km/h along a highway. A boy sitting near the driver of the RV throws a ball to another boy at the back end
alina1380 [7]

Answer:

Speed of the ball relative to the boys: 25 km/h

Speed of the ball relative to a stationary observer: 35 km/h

Explanation:

The RV is travelling at a velocity of

v_{RV}=+60 km/h

Here we have taken the direction of motion of the RV as positive direction.

The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is

v_B = -25 km/h

with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is

v = 25 km/h

Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:

v = +60 km/h + (-25 km/h) = +35 km/h

So, he/she measures a speed of 35 km/h.

5 0
2 years ago
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

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\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

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Answer:

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