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BabaBlast [244]
3 years ago
6

A ball is projected at an angle of 30° above the horizontal with a speed of 35 m/s.What will be its approximate horizontal range

across a level surface?
Physics
2 answers:
Lapatulllka [165]3 years ago
6 0

Answer:

Explanation:

I just need points

Oduvanchick [21]3 years ago
6 0

Answer:

24.89degree angle

Explanation:

Thats what i think because its my opinion it took me a while to answer your question but i wish you the best in luck with your degree angles

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I am pretty sure it is a cold front.
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4 years ago
______ 4. Justin was creating a chart comparing the potential energy between charged particles and the potential energy of magne
fredd [130]

Answer:

The correct option is;

A. The potential energy between both like charges and like poles increases as they move closer together

Explanation:

Here we have that when we move the like poles of two bar magnets close to each other, there is an increased resistance in the continuing motion, therefore for each extra gap closer achieved, there is an increase in potential energy

Similarly, when two like charges are brought closer together, the potential energy, or the energy available to push the two like charges apart increases charge as the as the charges are brought closer together

Therefore, the correct option is the potential energy between both like charges and like poles increases as they move closer together.

4 0
3 years ago
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DiKsa [7]

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4 0
3 years ago
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It's nighttime, and you ve dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90 m above the ed
yanalaym [24]

Answer:

The distance of the goggle from the edge is 5.30 m

Explanation:

Given:

The depth of pool (d) = 3.2 m

let 'i' be the angle of incidence

thus,

i = tan^{-1}(\frac{2.2}{0.90})

i = 67.75°

Now, Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

where,

r is the angle of refraction

n₁ is the refractive index of medium 1 = 1 for air

n₂ is the refractive index of medium 1 = 1.33 for water

now,

1 × sin 67.75° = 1.33 × sin(r)

or

r = 44.09°

Now,  

the distance of googles = 2.2 + d×tan(r)  = 2.2 + (3.2 × tan(44.09°) = 5.30 m

Hence, <u>the distance of the goggle from the edge is 5.30 m</u>

5 0
3 years ago
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