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levacccp [35]
3 years ago
9

A blue ball is thrown upward with an initial speed of 21.8 m/s, from a height of 0.9 meters above the ground. 2.7 seconds after

the blue ball is thrown, a red ball is thrown down with an initial speed of 10.4 m/s from a height of 26.6 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
Physics
1 answer:
worty [1.4K]3 years ago
3 0
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t, 
t = 2.25 seconds

Thus, the two balls would be at the same height after 2.25 seconds.
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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
What are the two most important properties of a telescope?
tigry1 [53]
1.Light-collecting area
2.Angular resolution
8 0
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If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa
vovikov84 [41]
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
  
          F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation. 
7 0
3 years ago
What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?
choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

Work = Force \times displacement

here we know that

Force = 8 N

displacement = 2 m

Now work done is given as

W = 8\times 2

W = 16 J

so it will do 16 J work to move the box

3 0
3 years ago
A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p
Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0
3 years ago
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