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Bess [88]
2 years ago
6

Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a gr

eater displacement?
Physics
2 answers:
MissTica2 years ago
5 0

Answer:

The distance of one lap on this track from the start line to the finish line is 400 meters. Two laps around the track is 800 meters, half a lap is 200 meters, and so on.

    Remember, displacement is the direction from the starting point and the length of a straight line from the starting point to the ending point. If a runner was going to run 400 meters (one lap around the track), they would start and stop at the same place on the track. Therefore, their displacement would be 0.

 

Explanation:

hope this helped

Ipatiy [6.2K]2 years ago
4 0

Answer:

\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has <u>no displacement.</u> If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has <u>50 meters of displacement</u>. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, <u>Gabi had the greater displacement.</u>

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Which expression of distance uses SI units? A. 30 miles B. 16 kilograms C. 24 feet D. 500 meters
Stella [2.4K]
Hi!

SI units are physical measurements which will be in the form of kilograms, second, kelvin, metres, etc.

Since kilograms measure the weight of an object, it is out. Miles and feet are not SI units, so they are also out. This only leaves one answer left!

Hopefully, this helps! =)
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3 years ago
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a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus
aivan3 [116]

Answer:

The modern instruments or we can say the different levels of telescopes are used to explore and study the distant galaxies. i.e the Hubble telescope is out there providing the data regarding the different properties of the celestial entities which in other case is not visible to the human naked eye.

Explanation:

  • Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.
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6 0
3 years ago
A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t
WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

3 0
2 years ago
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Ivanshal [37]

Answer:

Part a)

f_w = f_g = 4.57 \times 10^{14} Hz

Part b)

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\lambda_g = 437.3 nm

Part c)

v_w = 2.25 \times 10^8 m/s

v_g = 2.0 \times 10^8 m/s

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

f = \frac{v}{\lambda}

f = \frac{3 \times 10^8}{656 \times 10^{-9}}

f = 4.57 \times 10^{14} Hz

Part b)

As we know that the refractive index of water is given as

\mu_w = 4/3

so the wavelength in the water medium is given as

\lambda_w = \frac{\lambda}{\mu_w}

\lambda_w = \frac{656 nm}{4/3}

\lambda_w = 492 nm

Similarly the refractive index of glass is given as

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so the wavelength in the glass medium is given as

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\lambda_g = \frac{656 nm}{3/2}

\lambda_g = 437.3 nm

Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

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v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

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Answer:

Disagree

Explanation:

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