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Sever21 [200]
1 year ago
7

A shell is launched with an initial velocity at an angle of 40.0° above horizontal

Physics
1 answer:
olasank [31]1 year ago
7 0

Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40   = time in the air =<u> 1958.11 / x</u>

x sin 40 = vertical velocity

 find when shell vertical velocity = 0 (this is max height....1/2 way through its flight)  , the time when it hits the ground will be twice this...  

   0 =  x sin 40 - 9.8 t

  t =  x sin40 / 9.8        time in the air is twice this = .13118<u> x</u>

<u />

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s

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A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h
MrRissso [65]

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

                           t = d / v

Substituting the given values in the above equation,

                           t = 9.2 / 10

                             = 0.92 s

Hence, she ran for, t = 0.92 s

8 0
2 years ago
Air resistance is an example of what type of friction?
Sav [38]
Well, Air resistance is a special type of friction (you cannot classify it in other categories). That force of air-resistance is often observed to oppose the motion of the object,( like every other frictional forces)

Hope this helps!
4 0
3 years ago
How does a car get energy?
Lorico [155]

Answer:

Explanation:

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6 0
2 years ago
Read 2 more answers
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start
galina1969 [7]

Answer:

it moves 25 inches.

Explanation:

the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.

3 0
3 years ago
Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

5 0
3 years ago
Read 2 more answers
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