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3 years ago

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Physics

1 answer:

NikAS [45]3 years ago

7 0

D, Metamorphism
-Metamorphic rocks are any rocks that change into another rock when subjected to high heat and pressure.

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The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s

alex41 [277]

Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2

8 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

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3 years ago

A suitcase (mass m = 18 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures

Sindrei [870]

Answer:

P=2736 Pa

Explanation:

According to Newton we have that:

∑ $F=m*a\\$

A force is exerted by the elevator to the suitcase, according to 3th Newton's law an equal force but in the opposite direction will appeared on the suitcase, that is:

∑ $F=m*g+m*a=m*(g+a)$

$F=205.2N$

We know that the pressure is given by:

$P=\frac{F}{A}\\P=\frac{205.2N}{(0.50m)*(0.15m)}\\P=2736N/m^2=2736Pa$

6 0

3 years ago

3. J.J. Thomson discovered the electron in 1897. In 1904, he proposed a model

OLga [1]

Answer:

Maybe this will help?

Explanation:

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3 years ago

The diagram represents several forms of electromagnetic energy.

Naily [24]

Hello there.

The diagram represents several forms of electromagnetic energy.

Which feature best distinguishes one form of electromagnetic energy from another?

Answer: wavelength

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3 years ago