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LuckyWell [14K]
3 years ago
10

A ball is dropped off of a tall building and falls for 2 seconds before landing on a balcony. a rock is then dropped from the to

p of the building and falls for 4 seconds before landing on the ground. how does the final speed (meaning the speed it had just before landing) of the rock compare to the final speed of the ball?
Physics
1 answer:
koban [17]3 years ago
5 0
<span>Using the formula Final speed = initial speed + gravity * time, we can compare the results.

Being gravity approximately 9.8 m/s^2, and being both the rock and the ball dropped, we can assume that the initial speed of both is zero, so:

Ball: 9.8m/s^2 * 2s = 18.6 m/s
Rock: 9.8 m/s^2 * 4 = 37.2 m/s

The speed of the rock is the double of the speed of the ball at the moment just before landing, because it has been dropped in a greater height than the ball, allowing the rock to reach a greater speed than the ball.</span>
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So if n= 5 it can take 0,1,2,3,4

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for l=0 , m = 0 total = 1

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for l = 2, m=-2,-1,0,1,2 total = 5

5+3+1 = 9

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\sqrt(6) \times h'

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4 years ago
What component of earth’s atmosphere exists entirely as a result of photosynthesis?
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Explanation:

We must remember that the density of the substance is defined as the relationship between mass over volume. That is expressed in the following equation:

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Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:
FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle (v), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

v = R\cdot \omega (1)

Where:

R - Radius, measured in kilometers.

\omega - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that \omega \approx 5235.988\,\frac{rad}{s} and R = 1\times 10^{-4}\,km, then the linear velocity is:

v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)

v = 0.524\,\frac{km}{s}

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

\omega = \frac{2\pi}{T} (2)

Where T is the period, measured in seconds.

If we know that T = 31536000\,s, then the angular speed of the Earth is:

\omega = \frac{2\pi}{31536000\,s}

\omega = 1.992\times 10^{-7}\,\frac{rad}{s}

Now, we determine the linear speed:

v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)

v = 29.88\,\frac{km}{s}

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

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