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3 years ago

A student has a displacement of 304 m north in 180 s. What was the student's average velocity?

Physics

1 answer:

DanielleElmas [232]3 years ago

6 0

Answer:

v = 1.69 m/s

Explanation:

Given that,

Displacement of the student is 304 m due North and it takes 180 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

$v=\dfrac{304\ m}{180\ s}\\\\v=1.69\ m/s$

Hence, the student's average velocity is 1.69 m/s.

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882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91

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2 years ago

horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes

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Answer:

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Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

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The distance it sags is $d_s = 1m$

The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline

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And this can be mathematically represented as

$\sum F_y = 0$

To obtain $\theta$ we apply SOHCAHTOH Rule

So $Tan \theta = \frac{opp}{adj}$

$\theta = tan^{-1} [\frac{opp}{adj} ]$

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2 years ago

Read 2 more answers

A truck is moving at a constant speed of 50 m/s. A boy riding in the back of the truck throws a newspaper out the back of the tr

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Answer:

70m/s

Explanation:

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3 years ago

Please help me answer this

ivann1987 [24]

Answer:

A is the correct answer

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2 years ago

NASA communicates with the Space Shuttle and International Space Station using Ku-band microwave radio. Suppose NASA transmits a

jeyben [28]

Answer:

λ₁ = 2.50 10⁻² m, λ₂ = 1.66 10⁻² m

Explanation:

Microwave communication is very efficient because it does not have atmospheric interference, for which it is widely used and has been regulated to avoid interference, the ku band is in the range between 12 and 18 GHz.

Let's calculate the wavelength for the two extreme frequencies of this band

wavelength and frequency are related

c = λ f

λ = c / f

f₁ = 12 GHz = 12 10⁹ Hz

λ₁ = 3 10⁸ /12 10⁹

λ₁ = 2.50 10⁻² m

f₂ = 18 GHz = 18 10⁹ Hz

λ₂ = 3 10⁸ /18 10⁹

λ₂ = 1.66 10⁻² m

Unfortunately in your exercise the specific frequency is not fired, for significant figures they must be the same number as the figures of the frequency, in general the frequency has 3 or 4 significant figures

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3 years ago