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Harman [31]
3 years ago
8

A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

building under construction. A 70 kg construction worker stands at the far end of the beam.
What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?
Physics
1 answer:
icang [17]3 years ago
4 0

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m

Torque due to person

T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m

Now for total torque

T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m

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How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air
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y = 75.9 m

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y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

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2 years ago
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Paul [167]

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2 years ago
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​
WARRIOR [948]

\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

Explanation:

Given:

\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}

\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}

The cross product \textbf{A}×\textbf{B} is given by

\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|

=  \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}

= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

5 0
3 years ago
10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????
KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5×10^{-6} m.

              Tension = 95.6N, Δx = 15.4×10^{-5} m

Explanation: A speed of wave on a string under a tension force can be calculated as:

|v| = \sqrt{\frac{F_{T}}{\mu} }

F_{T} is tension force (N)

μ is linear density (kg/m)

Determining velocity:

|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }

|v| = \sqrt{0.00874 }

|v| = 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

\Delta x = |v|.t

\Delta x = 9.35.10^{-2}*1.23.10^{-3}

Δx = 11.5×10^{-6}

With tension of 47.8N, a pulse will travel Δx = 11.5×10^{-6}  m.

Doubling Tension:

|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }

|v| = \sqrt{2.0.00874 }

|v| = \sqrt{0.01568}

|v| = 0.1252 m/s

Displacement for same time:

\Delta x = |v|.t

\Delta x = 12.52.10^{-2}*1.23.10^{-3}

\Delta x = 15.4×10^{-5}

With doubled tension, it travels \Delta x = 15.4×10^{-5} m

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3 years ago
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