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svp [43]
3 years ago
9

Write any two uses of a wedge.​

Physics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer: two uses of a wedge can be to tighten things as well as to lift things.

Explanation: because that is some things wedges (a mechanical tool) can do.

You might be interested in
A flight distance from philadelphia to cancun is 1469 miles.the estimated flight time is 3.25 hours.What is the avarage speeede
Jlenok [28]

Answer:

1469 miles /3.25 hours = 452 miles per hours

Explanation:

divide distance by the time it takes to get there

4 0
3 years ago
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.
olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

U_{g,1} + K_{x,1} + K_{y,1} =  U_{g,2} + K_{x,2} + K_{y,2}

Where:

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

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

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

Now, the system is expanded and simplified:

m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0

g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})

y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}

Where:

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

v_{y,1}, v_{y,2} - Initial and final vertical speed of the arrow, measured in meters.

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

The initial vertical speed of the arrow is:

v_{y,1} = v_{1}\cdot \sin \theta

Where:

v_{1} - Magnitude of the initial velocity, measured in meters per second.

\theta - Initial angle, measured in sexagesimal degrees.

If v_{1} = 82\,\frac{m}{s} and \theta = 24^{\circ}, the initial vertical speed is:

v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}

v_{y,1} \approx 33.352\,\frac{m}{s}

If g = 9.807\,\frac{m}{s^{2}}, v_{y,1} \approx 33.352\,\frac{m}{s} and v_{y,2} = 0\,\frac{m}{s}, the maximum height of the first arrow is:

y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }

y_{2} - y_{1} = 56.712\,m

Second arrow

U_{g,1} + K_{y,1} =  U_{g,3} + K_{y,3}

Where:

U_{g,1}, U_{g,3} - Initial and final gravitational potential energy, measured in joules.

K_{y,1}, K_{y,3} - Initial and final vertical translational kinetic energy, measured in joules.

m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0

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

y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}

If g = 9.807\,\frac{m}{s^{2}}, v_{y,1} = 82\,\frac{m}{s} and v_{y,3} = 0\,\frac{m}{s}, the maximum height of the first arrow is:

y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }

y_{3} - y_{1} = 342.816\,m

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

E = U + K_{x}

The expression is now expanded:

E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}

Where v_{x} is the horizontal speed of the arrow, measured in meters per second.

v_{x} = v_{1}\cdot \cos \theta

If v_{1} = 82\,\frac{m}{s} and \theta = 24^{\circ}, the horizontal speed is:

v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}

v_{x} \approx 74.911\,\frac{m}{s}

If m = 0.06\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{max} = 56.712\,m and v_{x} \approx 74.911\,\frac{m}{s}, the total mechanical energy is:

E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}

E = 201.720\,J

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

E = m\cdot g \cdot y_{max}

m = 0.06\,kg, g = 9.807\,\frac{m}{s^{2}} and y_{max} = 342.816\,m

E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)

E = 201.720\,J

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

7 0
3 years ago
A constant tone is being applied to a speaker. The voltage across the speaker is 5 volts. The voltage across the speaker is incr
lubasha [3.4K]

Answer:

decibel gain is 10 db

Explanation:

given data

voltage v1 =  5 volts

voltage v2 = 15 volts

to find out

decibel gain

solution

we have given that 5 volt voltage that is increase to 15 volts

so

we know here decibel gain formula that is

decibel gain = 20 log(v2/v1)   ....................1

put here value of v1 and v2  in equation 1

decibel gain = 20 log(15/5)

decibel gain = 20 log3

decibel gain = 20 × 0.4771

decibel gain = 9.54 db = approx 10 db

so decibel gain is 10 db

5 0
4 years ago
Read 2 more answers
HELP PLEASE !! how are waves, engery, and matter related??
Makovka662 [10]

Waves transport energy along a medium without transporting matter. The amount of energy carried by a wave is related to the amplitude of the wave. Thus, the higher the wave is from the resting line, the more energy is put in and vice-versa.

7 0
4 years ago
Which statement represents how charges interact?
Serhud [2]
Positive charges are attracted to negative charges. From magnetism to interactions between molecules, objects with opposite charges are attracted to each other, and objects with the same charge repel one another. 
3 0
3 years ago
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