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2 years ago

Your favorite taco stand is 800 m away, and closes in 15 minutes. How fast do you have to run to make it on

Physics

2 answers:

Dimas [21]2 years ago

8 0

You must run at least 53.3333333 meters a minute.

ValentinkaMS [17]2 years ago

6 0

To find the rate that you need to run, you should do distance/time

So, 800/15=53.33. You have to run at 53.33meters per minute to make it on time

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Which of the following statements is untrue in regard to sound traveling in air?

maksim [4K]

Air never stops? IDK

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2 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago

Can someone answer this for me? it’s due today and i need it done asap! i will give brainlist!!

Setler79 [48]

Answer:

1. The sound waves are longitudinal because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.

2. A pulse or a wave is introduced into a slinky when a person holds the first coil and gives it a back-and-forth motion. This creates a disturbance within the medium; this disturbance subsequently travels from coil to coil, transporting energy as it moves.

Explanation:

5 0

2 years ago

Read 2 more answers

Halleys comet has period of 75.3 years. Using Kepler’s third law, find it’s semimajor axis expressed in astronomical units?

natta225 [31]

Answer: 17.83 AU

Explanation:

According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. 

$T^{2}\propto a^{3}$ (1)

Talking in general, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet) orbiting a greater body in space with the size $a$ of its orbit.

However, if $T$ is measured in years, and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes: