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Anestetic [448]
3 years ago
15

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is

the number of 3-point questions and y is the number of 5-point questions, the system shown represents this situation. x + y = 24 3x + 5y = 100 What does the solution of this system indicate about the questions on the test?
A. The test contains 4 three-point questions and 20 five-point questions.
b. The test contains 10 three-point questions and 14 five-point questions.
c.The test contains 14 three-point questions and 10 five-point questions.
d. The test contains 20 three-point questions and 8 five-point questions.
Mathematics
1 answer:
loris [4]3 years ago
6 0
B.
So when you solve the system of equations you get that y=14 and x=10. When you plug those into the original equations you get true statements. 
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K_{1}+U_{g,1} = K_{2}+U_{g,2} (1)

Where:

K_{1}, K_{2} - Initial and final translational kinetic energies, measured in joules.

U_{g,1}, U_{g,2} - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:

\frac{1}{2}\cdot m\cdot v_{2}^{2}+m\cdot g\cdot y_{2}= \frac{1}{2}\cdot m\cdot v_{1}^{2}+m\cdot g\cdot y_{1} (2)

Where:

m - Mass of the ball, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

v_{1}, v_{2} - Initial and final speed of the ball, measured in meters per second.

y_{1}, y_{2} - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:

v_{2}^{2}+2\cdot g\cdot y_{2} = v_{1}^{2}+2\cdot g\cdot y_{1}

v_{1}^{2}-v_{2}^{2}+2\cdot g \cdot y_{1}=2\cdot g\cdot y_{2}

y_{2} = y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} (3)

If we know that y_{1} = 30\,m, v_{1} = 14\,\frac{m}{s}, v_{2} = 0\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, the maximum height that the ball will reach is:

y_{2} = 30\,m + \frac{\left(14\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

y_{2} = 39.993\,m

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:

t = \frac{v_{2}-v_{1}}{-g} (4)

If we know that v_{1} = 14\,\frac{m}{s}, v_{2} = 0\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, then:

t = \frac{0\,\frac{m}{s}-14\,\frac{m}{s}  }{-9.807\,\frac{m}{s^{2}} }

t = 1.428\,s

The ball will take 1.428 seconds to reach its maximum height.

6 0
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