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Alja [10]
3 years ago
11

It is possible to hang from a bare power line and not get electrocuted as long as you are not touching the ground, or any conduc

tor that is touching the ground.
Physics
1 answer:
NNADVOKAT [17]3 years ago
3 0
Yes i believe so. depending on the covering of the line and other environmental confounding variables 
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A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area
Furkat [3]

Answer:

DL = 1.5*10^-4[m]

Explanation:

First we will determine the initial values of the problem, in this way we have:

F = 60000[N]

L = 4 [m]

A = 0.008 [m^2]

DL = distance of the beam compressed along its length [m]

With the following equation we can find DL

\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m]

Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.

A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D =  \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m]

3 0
4 years ago
A hiker walks 15km due east then heads due north for 8km . What is the direction of the resultant vector?
4vir4ik [10]

his displacements are given as

d_1 = 15 km East

d_2 = 8 km North

now we can find the direction of displacement by using the concept

\theta = tan^{-1}\frac{\deta y}{\delta x}

\theta = tan^{-1}\frac{8}{15}

\theta = 28.1 degree

magnitude of displacement is given as

d = \sqrt{x^2 + y^2}

d = \sqrt{15^2 + 8^2}

d= 17 km

<em>so the displacement is 17 km at angle 28.1 degree North of East</em>

8 0
3 years ago
I really don’t know what it is to me they r all the same
Inessa05 [86]
The moon lacks an atmosphere compared to the Earth. Hope this helps!
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3 years ago
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Explanation:

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Answer:

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