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Anastasy [175]
3 years ago
14

A soccer player hits a ball with a velocity of 22 m/s at an angle of 36.9 above the horizontal. Air resistance can be ignored. a

. b. c. (3) What are the x and y components of the balls initial velocity? (3) How high does the ball go? (4) How long (time) does it take to get to the maximum height?
Physics
1 answer:
TiliK225 [7]3 years ago
4 0
Answers:

a. Vx=17.6m/s
Vy=13.2m/s

b. 8.90m

c. 1.80s




Vi=22m/s

Theta=36.9deg

Vx=Vi*cos(theta)

Vy=Vi*sin(theta)

h(Max)=Vi^2*sin(theta)^2/2g

t(max height)= Vy/g

Vx=22*cos(36.9)=17.6m/s

Vy=22*sin(36.9)=13.2m/s

h(Max)=22^2*sin(36.9)^2/2(9.8)=8.90m

t(Max height)=17.6/9.8=1.80s
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For this case we have that by definition, the kinetic energy is given by the following formula:

k= \frac {1} {2} * m * v ^ 2

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

m = 100 \ kg\\v = 9 \frac {m} {s}

Substituting the values we have:

k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050

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Answer:

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
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Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

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From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

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Explanation:

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