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nexus9112 [7]
2 years ago
15

Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

d why
Physics
1 answer:
blsea [12.9K]2 years ago
7 0
They are both right because you can note both things, I mean Raphael and Lucinda, both has a right statement or explanation about the wave. Wave by nothing is both for its wavelength and for its frequency. So Raphael and Lucinda are both correct because you can note both wavelength and frequency.
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If you take any pitch on the keyboard, the next occurrence of the same letter name going towards the left (down) will vibrate:
NikAS [45]

Answer:

A. Twice as slow

Explanation:

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1 year ago
A car drives around a racetrack for 30 seconds. what do you need to know to calculate the average velocity of the car?
boyakko [2]
The time is given, and you want to find the average velocity. To do this, you need to know the distance covered by the driver around the racetrack in that 30 seconds. You divide this by the time, then you will obtain the average velocity in units of, say meters per second.
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Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b
Andrews [41]
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6 0
2 years ago
An object has a position given by r = [2.0 m + (2.00 m/s)t] i + [3.0 m − (1.00 m/s^2)t^2] j, where quantities are in SI units. W
lidiya [134]

Answer: 1 m/s

Explanation:

We have an object whose position r is given by a vector, where the components X and Y are identified by the unit vectors i and j (where each unit vector is defined to have a magnitude of exactly one):

r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j

On the other hand, velocity is defined as the variation of the position in time:

V=\frac{dr}{dt}

This means we have to derive r:

\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j

\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} t) j This is the velocity vector

And when t=2s the velocity vector is:

\frac{dr}{dt}=(2 m/s) i - (\frac{1}{2} m/s^{2} (2 s)) j

\frac{dr}{dt}=2 m/s i - 1m/s j This is the velocity vector at 2 seconds

However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed S:

S=\sqrt {-1 m/s j + 2 m/s i}

S=\sqrt {1 m/s}

Finally:

S=1 m/s This is the speed of the object at 2 seconds

6 0
1 year ago
A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take
Ad libitum [116K]

Answer:

F=(3i+3.6j)\ N

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, u=(1i+0j)\ m/s

After 8 seconds, final velocity of the puck, v=(6i+6j)\ m/s

Let the x and y component of force is given by F_x\ and\ F_y.

x component of force is given by :

F_x=m\times \dfrac{v-u}{t}

F_x=4.8\times \dfrac{6-1}{8}

F_x=3\ N

y component of force is given by :

F_y=m\times \dfrac{v-u}{t}

F_y=4.8\times \dfrac{6-0}{8}

F_y=3.6\ N

So, the component of the force is F=(3i+3.6j)\ N. Hence, this is the required solution.

7 0
2 years ago
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