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madreJ [45]
3 years ago
12

A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula

te the tension.​
Physics
1 answer:
Snezhnost [94]3 years ago
7 0

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

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zubka84 [21]
The x-component of Anne's path:
80 x cos(33)
= 67.1 meters
y-component of Anne's path:
80 x sin(33)
= 43.6 meters
To draw he diagram, you draw a horizontal line starting at point 1 to the right representing 67.1 meters. At the end of this line, draw a vertical line upwards representing 43.6 meters.
Connect the start of the horizontal line at point 1 to the end of the vertical line with an angle of 33° between them.
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3 years ago
A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right.
lisov135 [29]

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

\Sigma F = F - T = m_{J}\cdot a (1)

Glider

\Sigma F = T = m_{G}\cdot a (2)

Where:

F - Thrust of jet engines, measured in newtons.

T - Tension in the cable connecting the jet and glider, measured in newtons.

m_{G}, m_{J} - Masses of the glider and the jet, measured in kilograms.

a - Acceleration of the glider-jet system, measured in meters per square second.

If we know that m_{G} = 260\,kg, m_{J} = 1,940\,kg and a = 2.20\,\frac{m}{s^{2}}, then the solution of this system of equations:

By (2):

T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)

T = 572\,N

By (1):

F = T+m_{J}\cdot a

F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)

F = 4840\,N

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

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3 years ago
A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0° above the horizontal. it strike
Anni [7]
<span>x = 129.9 m y = 30.9 m First, let's calculate the horizontal and vertical velocities involved h = 50.0cos(30) = 43.30127 m/s v = 50.0sin(30) = 25 m/s The horizontal distance is simply the horizontal velocity multiplied by the time, so 43.30127 m/s * 3 s = 129.9 m So the horizontal distance traveled is 129.9 m, so x = 129.9 m The vertical distance needs to take into account gravity which provides an acceleration of -9.8 m/s^2, so we get d = 25 m/s * 3s - 0.5*9.8 m/s^2 * (3 s)^2 d = 75 m - 4.9 m/s^2 * 9 s^2 d = 75 m - 44.1 m d = 30.9 m So the vertical distance traveled is 30.9 m, so y = 30.9 m</span>
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4 years ago
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lisov135 [29]

If you stand up in a big room and echo, your voice will echo from the walls. As long as the room is empty. Since the speed of sound is constant, depending on air density, the more humid the air the faster and farther sound travels. The speed of sound is constant, you could measure the time it takes for your voice to echo off the walls. The same thing happens with Doppler radar, but it’s not voice, it has higher frequency signals.<span> </span>

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3 years ago
Read 2 more answers
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sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

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so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0
3 years ago
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