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xxMikexx [17]
2 years ago
6

A punted football is observed to have velocity components vhorizontal = 15 m/s to the right and vvertical = 1.25 m/s directed do

wnward at a height h = 1.3 m above the ground. Use a Cartesian coordinate system with the origin located on the ground at the position the football was punted and assume it encountered no air resistance. Determine the football's initial horizontal velocity magnitude v h in terms ofw roe in. Wermeal, g, and h.
Physics
1 answer:
enot [183]2 years ago
5 0

Answer:

v₀ₓ = 15 m / s,  v_{oy} = 5.2 m / s

v = 15.87 m / s ,   θ = 19.1

Explanation:

This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed

                      Vy² =v_{oy}² - 2 g y

                      v_{oy}² = v_{y}² + 2 g y

                       v_{oy} = √ (v_{y}² + 2 gy

Let's calculate

                    v_{oy} = √ (1.25² + 2 9.8 1.3)

                    v_{oy} = √ (27.04)

                    v_{oy} = 5.2 m / s

 The initial speed can be calculated by the initial speed

                   v = √ v₀ₓ² + v_{oy}²

                   v = RA (15² + 5.2²)

                   v = 15.87 m / s

We look for the angle with trigonometry

                 tan θ = voy / vox

                 θ = tan⁻¹ I'm going / vox

                θ = tan⁻¹ 5.2 / 15

                θ = 19.1

The answer is

              v₀ₓ = 15 m / s

              v_{oy} = 5.2 m / s

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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
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Answer:1.63 m

Explanation:

Given

mass of block m=0.5 kg

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Amount of work done W=4 J

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is s\sin \theta

Change in Potential Energy =mg(\sin \theta -0)

\Delta P.E.=0.5\times 9.8\times s\sin 30

4=0.5\times 9.8\times s\sin 30

s=\frac{4}{2.45}      

s=1.63 m

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2 years ago
Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E
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Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

H = \vec d

work done is given as

W = \vec F . \vec d

Potential energy is given as

PE_g = mgh

force due to gravity is given as

\vec F_g = mg

now here if we plug in the value of distance and force in the formula of work done then we will have

W = (mg)(h)

so here we got

W = PE_g

so we can concluded that

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2 years ago
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A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer
matrenka [14]

Answer:

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F = m \frac{(v_f - v_o)}{t}

let's calculate

         F = 1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v = \sqrt{2g y_o}

calculate  

         v = \sqrt{2 \ 9.8 \ 10}

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

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4 0
1 year ago
An object on a number line moved from x = 15 cm to x = 165 cm and then
olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

v_{avg}=\frac{x_{all}}{t}

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0
2 years ago
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If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee
Alexus [3.1K]

Answer:

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Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0
2 years ago
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