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shepuryov [24]
3 years ago
8

A basketball is thrown straight up in the air. At its peak, it is 0.02 km high and has a velocity of 0 m/s. What is its final ve

locity when it hits the ground?
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

20 m/s

Explanation:

Given:

Δy = 0.02 km = 20 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (20 m)

v = 19.8 m/s

Rounded to one significant figure, the final velocity is 20 m/s towards the ground.

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Answer:

When they are connected in series

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Where  V is the normal household voltage which is constant for both bulbs

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substituting values

        R_1  =  \frac{V^2}{100}

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       R_2  =  \frac{V^2}{50}

From the above calculation we see that

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This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

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     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

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