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amm1812
3 years ago
7

1. Transverse waves _____.

Physics
2 answers:
poizon [28]3 years ago
8 0
D. have particle moment perpendicular to the direction of the energy.
kumpel [21]3 years ago
7 0
The answer is D to you question 
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The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands
Ghella [55]

Answer:

Pressure on both feet will be 500N/m^2  

Explanation:

Weight of the person F = 500 N

Area of foot print A=0.5m^2

Area of both the foot A=2\times 0.5=1m^2

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure P=\frac{F}{A}

P=\frac{500}{1}=500N/m^2

So the pressure on both feet will be 500N/m^2 when person stands on both feet.

7 0
3 years ago
4. Which scientist proposed the theory that the Moon formed from fission? (5 points)
NeTakaya
The answer would be B, George Darwin :)
7 0
3 years ago
What happens to the current as we increase the amount of stepping of our transformer? Does this help explain why the primary was
matrenka [14]

Answer:

Current will decrease.

Explanation:

When we increase the number of stepping in transformer, the voltage will increase as its is directly proportional to the number of turn of stepping. Thus as the voltage will increase, current will decrease. As per the equation of ideal transformer,   E1 / E2 = I2 / I1

E1 and E2 are the voltages in primary and secondary winding and I1 and I2 are the current.

As the number of turns will be increased more inevitable losses will be generated that dissipates heat thus warming the primary.

Though the conservation of energy is obeyed but losses occur in this scenario hence step-up transformers cannot be used to create free energy.

7 0
3 years ago
Physical quantity are measurable quantities
Vilka [71]

Answer:

yeah physical quantities are the quantities which can be meaured

7 0
3 years ago
At what distance along the z-axis is the electric field strength a maximum?
Lesechka [4]
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is 

<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>

<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>

<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>

<span>so the equation becomes </span>

<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>

<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>

<span>Integrating around the ring you get </span>

<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>

<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>

<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>

<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>

<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>

<span>to find the maxima set this = 0, giving </span>

<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>

<span>mult both side by (x^2 + R^2)^2.5 to get </span>

<span>(x^2 + R^2) - 3*x^2 = 0 </span>

<span>-2*x^2 + R^2 = 0 </span>

<span>-2*x^2 = -R^2 </span>

<span>x = (+/-)R/sqrt(2) </span>
4 0
3 years ago
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