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RSB [31]
2 years ago
10

Identrying the speed of the Cart

Physics
2 answers:
FrozenT [24]2 years ago
7 0

The speed of the cart after 3 seconds of Low fan speed is 55 cm/s.

The speed of the cart after 5 seconds of Medium fan speed is 120 cm/s.

The speed of the cart after 2 seconds of High fan speed is 63 cm/s.

<h3>What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

The graph in the attachment may be used to estimate the cart speed for various fan speeds.

Blue letters indicate the low fan speed. When the line drawn from x hits the blue line, the elapsed time is indicated on the x-axis cart speed after three seconds.

This is equivalent to 55 cm/s.Yellow markings indicate the medium fan speed. After 5 seconds, the line drawn from the point marked 5 on the x-axis crosses the yellow line, marking the time that has passed.

The intersection of the yellow line with the line drawn from the point marked 5 on the x-axis indicates the cart speed five seconds later. The y-axis measurement is roughly 120 cm/s.

Green markings indicate the high fan speed. On the x-axis is the elapsed time. Cart speed at the place where the green line and the line drawn from the x-point axis 2 connect. The y-axis reading is roughly 63 cm/s.

Hence, after the specified seconds, the cart's estimated speeds at low, medium, and high speeds are 55 cm/s, 120 cm/s, and 63 cm/s, respectively.

To learn more about the speed refer to the link;

brainly.com/question/7359669

#SPJ1

aleksklad [387]2 years ago
6 0

The speed of the cart after 3 seconds of Low fan speed is equal to 54 cm/s.

<h3>How to calculate the speed?</h3>

Mathematically, speed can be calculated by using this formula;

Speed = distance/time

At Low fan speed after 3 seconds, the distance covered is 162 cm:

Speed = 162/3

Speed = 54 cm/s.

At Medium fan speed after 5 seconds, the distance covered is 600 cm:

Speed = 600/5

Speed = 120 cm/s.

At High fan speed after 2 seconds, the distance covered is 128 cm:

Speed = 128/2

Speed = 64 cm/s.

Read more on speed here: brainly.com/question/17350470

#SPJ1

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Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

Thus the angular separation becomes

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6 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

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f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

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Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

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3 years ago
If the Moon had twice as much mass and still orbits Earth at the same distance, ocean bulges on Earth would be
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Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

<h3>What is ocean bludge?</h3>

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

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maxonik [38]

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