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Free_Kalibri [48]
3 years ago
7

The table shows information about for objects resting at the top of a hill. Which object is on the tallest hill?

Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
6 0
The formula for Potential Energy is as follows:

Potential Energy = mass x gravity x height

Begin by inserting any known values into the equation above:

W: 980 = (50)(9.8)(h)
X: 1029 = (35)(9.8)(h)
Y: 1519 = (62)(9.8)(h)
Z: 1176 = (24)(9.8)(h)

Then simplify the right side of each equation:

W: 980 = 490h
X: 1029 = 343h
Y: 1519 = 607.6h
Z: 1176 = 235.2h

Divide the coefficient of “h” from both sides of each equation:

W: (980/490) = h
X: (1029/343) = h
Y: (1519/607.6) = h
Z: (1176/235.2) = h

Simplify the left side of each equation:

W: 2 = h
X: 3 = h
Y: 2.5 = h
Z: 5 = h

Object Z has a height of 5. Object Z is sitting on the tallest hill.

I hope this helps!
Fudgin [204]3 years ago
5 0

Answer:

the answer is z

Explanation:

just took the test and got 100%

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3 years ago
If the distance of a galaxy is 2,000 Mpc, how many years back into the past are we looking when we observe this galaxy
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The age of the galaxy when we look back is 13.97 billion years.

The given parameters:

  • <em>distance of the galaxy, x = 2,000 Mpc</em>

According Hubble's law the age of the universe is calculated as follows;

v = H₀x

where;

H₀ = 70 km/s/Mpc

T = \frac{x}{V} \\\\T = \frac{x}{xH_0} \\\\T = \frac{1}{H_0} \\\\T = \frac{1}{70 \ km/s/Mpc} \\\\T = \frac{1 \ sec}{70 \times 3.24 \times 10^{-20} } \\\\T = 4.41 \times 10^{17} \ sec\\\\T = \frac{4.41 \times 10^{17} \ sec\  \times \ years}{3600 \ s \ \times\  24\ h\  \times \ 365.25 \ days} \\\\T = 1.397  \times 10^{10} \ years\\\\T = 13.97 \ billion \ years

Thus, the age of the galaxy when we look back is 13.97 billion years.

Learn more about Hubble's law here: brainly.com/question/19819028

8 0
2 years ago
What concept or principle best explains why
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Answer:

d. Newton's first law

I hope this helps you

5 0
3 years ago
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KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

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