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zheka24 [161]
3 years ago
9

Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal

ly and 2.5 mm, vertically below. If she long jumps from the edge of the left bank at 45 ∘∘ with the speed calculated in A, how long, or short, of the opposite bank will she land?
Physics
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

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Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

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Using Pythagorean theorem

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D = \sqrt{(400)^2+(180)^2}

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Answer:

Explanation:

1 )

Here

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For angular position of θ1

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In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.
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The central force acting on the electron as it revolves in a circular orbit is 9.52 \times 10^{-8} \ N.

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<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

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