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lora16 [44]
3 years ago
14

Through what media does mechanical waves move

Physics
1 answer:
Aleks [24]3 years ago
6 0
A mechanical wave moves through all matter
You might be interested in
A 12,000 kg railroad car is traveling at 2 m/s when it strikes another 10,000 kg railroad car that is at rest. If the cars lock
Degger [83]

Answer:

The final speed of the railroad car

V= 1.14 \frac{m}{s}

Explanation:

v_{1}=2.1\frac{m}{s} \\m_{1}=12000kg\\v_{2}=0\frac{m}{s} \\m_{2}=10000kg \\v_{t}=?

m_{1}*v_{1}+m_{2}*v_{2}= (m_{1}+m_{2})*v_{t}\\v_{t}=\frac{m_{1}*v_{1}}{(m_{1}+m_{2})} \\v_{t}=\frac{12000kg*2.1\frac{m}{s} }{(12000+10000)kg} \\v_{t}=1.14 \frac{m}{s}

That's the final speed of the both railroad car

4 0
3 years ago
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
3 years ago
What specific space object is shown in the picture below?
PIT_PIT [208]
From context clues, I believe the correct answer is B) Red Dwarf!
3 0
3 years ago
Fluid clutches are used on equipment that is subjected to?
mamaluj [8]
Fluid clutches are used on equipment that is subjected to higher torque per unit volume. This type of clutch makes use of an incompressible fluid like oil in order to transfer the input of a movement from a pedal and cylinder to an actuating cylinder. Releasing the this clutch would stop the transfer of power but still allowing the engine to turn continuously. 
4 0
3 years ago
A javelin thrower standing at rest holds the center of the javelin behind her head, then accelerates it through a distance of 70
JulijaS [17]

Answer:

a = 580 m/s^2

Explanation:

Given:

- Distance for accelerated throw s_a = 70 cm

- Angle of throw Q = 30 degrees

- Distance traveled by the javelin in horizontal direction x(f) = 75 m

- Initial height of throw y(0) = 0

- Final height of the javelin y(f) = -2 m

Find:

What was the acceleration of the javelin during the throw? Assume that it has a constant acceleration.

Solution:

- Compute initial components of the velocity:

                                             V_x,i = V*cos(30)

                                             V_y,i = V*sin(30)

- Use second equation of motion in horizontal direction:

                                          x(f) = x(0) + V*cos(30)*t

                                            75 = 0 + V*cos(30)*t

                                              t = 75 /V*cos(30)

- Use equation of motion in vertical direction:

                                     y(f) = y(0) + V_y,i*t + 0.5*g*t^2

Subs the values:

                      -2 = 0 + V*sin(30)*75/V*cos(30) - 4.905*(75/Vcos(30))^2

                           -2 = 75*tan(30) - 4.905*(5625/V^2*cos^2(30))

                           V^2 = 4.905*5625 / (2 + 75*tan(30))*cos^2(30)

                                                V^2 = 812.0633

                                                 V = 28.5 m/s

- Use the third equation of motion in the interval of the throw:

                                            V^2 = U^2 + 2*a*s_a

                                               28.5^2 = 2*a*0.7

                                                a = 580 m/s^2

         

     

6 0
3 years ago
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