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4 years ago

13

A body is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150N.

What is the tension in each wire?

1 answer:

Maru [420]4 years ago

8 0

The aid of a free-body diagram is essential for solving this mechanics problem. The summation of horizontal and vertical forces in the system must be equal to zero since the body is in equilibrium.

Summation of x-forces:

T*cos(40)-T*cos(40) = 0

y forces:

2T*sin(40) - 150 = 0

T = 150/2*sin(40) = 100.66 N

Therefore, the tension in each wire would be 100.66 Newtons.

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A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

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if a ball on a string swung in a circles and string suddenly breaks.the ball will move into what direction?

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Answer: b

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Why does an iceberg have more HEAT than a cup of coffee?

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Answer:

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3 years ago

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

$\tau=F\times r$

$\tau=82.0\times0.150$

$\tau=12.3\ N-m$

Now, The angular acceleration

$\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}$

$\dfrac{d\omega}{dt}=20.73\ rad/s^2$

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

$\tau=I\times\dfrac{d\omega}{dt}$

$I=\dfrac{I}{\dfrac{d\omega}{dt}}$

$I=\dfrac{12.3}{20.73}$

$I= 0.593\ kg-m^2$

Hence, The moment of inertia of the wheel is 0.593 kg-m².

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