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OleMash [197]
3 years ago
15

What is the distance between a(-6,3),b(-2,4),c(-8,3)​

Physics
1 answer:
Schach [20]3 years ago
8 0

Answer:

Explanation:

Appears to be the vertexes of a triangle.

AB = √(-6 - (-2))² + (3 - 4)²) = √17

AC = √(-6 - (-8))² + (3 - 3)²) = 2

BC = √(-2 - (-8))² + (4 - 3)²) = √37

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You dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.15 m. Determine t
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Frequency = rate of sploosh = 2 per second  =  2 Hz.

Period = ( 1/frequency ) =  1/2  second

Speed = (wavelength) x (frequency) = (0.15m) x ( 2/sec) = 0.075 m/s .

5 0
3 years ago
Muscular Endurance is the ability of a muscle to continue to perform without fatigue.
Art [367]

Answer:

True.

Explanation:

Defenintion of Muscular Endurance:  

The ability of a muscle (or set of muscles) to perform a repeated action without tiring.

7 0
3 years ago
Which steps can be taken to translate the phrase “the height of a tree is increased by seven inches” into an algebraic expressio
ololo11 [35]

Answer:

Explanation:

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increaced by=+

3 0
2 years ago
Read 2 more answers
What's the difference between a direct relationship and a positive one?
borishaifa [10]
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5 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
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