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Alja [10]
2 years ago
14

3 What is the displacement of a satellite when it makes a complete round along its circular path?​

Physics
1 answer:
kirill [66]2 years ago
4 0

Answer:

0

Explanation:

The displacement is zero since it goes in a full circle and ends up where it started.

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A scientific hypothesis must be related to nature and be..
PSYCHO15rus [73]

Answer:

A scientific hypothesis must be tetable so it can become a scientific theory.

Explanation: I think

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3 years ago
A loaf of bread (volume 3100 cm3) with a density of 0.90 g/cm3 is crushed in the bottom of the grocery bag into a volume of 1240
andrew11 [14]

Answer:

2,25 g/cm3

Explanation:

Hi, you have to know one thing for this.. Density = mass/Volume,

When you have the loaf of bread with 3100 cm3 and a density of 0.90 g/cm3, the mass of that bread is 2790 g because of if you isolate the variable mass from the equation you get..  mass= density x volume

Later, have on account the mass never changes, so you crush the bread and the mass is the same.. so when you have the mashed bread.. you know that the mass is 2790 g and the volume of the bag is 1240 cm3, so you apply the main equation.... density=2790 g / 1240 cm3 , so density =  2,25 g/cm3

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3 years ago
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Answer:

ok

Explanation:

or zo0m meet??

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3 years ago
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Which best compares AC and DC?
Naily [24]
Yea im pretty sure its B
5 0
3 years ago
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While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th
Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = \dfrac{2 \pi r}{T}

v = \dfrac{2\times \pi \times 15}{23}

v = 4.10 m/s

b) The magnitude of the acceleration is:

a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma  

apparent weight =mg - ma  

\dfrac{apparent \ weight}{true \  weight} = \dfrac{(mg - ma)}{(mg)}

=1- \dfrac{1.12}{9.8}

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}

\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}

= 1 + \dfrac{1.12}{9.8} \\ \\

= 1.114 m/s²

6 0
3 years ago
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