Answer:
Hey :)
Explanation:
Work is a net force applied through a distance in order to displace an object, commonly abbreviated as W. A net force is the sum of all forces acting on an object. Work is mass times acceleration and distance so to find out the work you simply calculate the acceleration of the box being brought in. Next find the distance it was carried to get in the house. Then find out the mas of the box and finally multiply those sums together to get the amount of work put in to bring the package inside.
hope this helps :) xo
Options:
A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.
B. Star 1 is 100 times more distant than Star 2.
C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.
D. Star 1 is 10 times more distant than Star 2.
E. Star 1 is 100 times nearer than Star 2.
Answer:
D. Star 1 is 10 times more distant than star 2
Explanation:
For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.
Luminosity, L = 1/r²
Where r is the distance of the star to the earth
Since star 1 is dimmer in brightness than star 2 by a factor of 100,
L₁/L₂ = 1/100
i.e. L₁ = 1, L₂=100
L₁ = 1/r₁² ............(1)
1 = 1/r₁²
L₂ = 1/r₂²
100 = 1/r₂² .........(2)
divide equation (2) by equation (1)
100/1 = ( 1/r₂² )/ (1/r₁²)
100 = (r₁/r₂)²
r₁/r₂ = √100
r₁/r₂ = 10
r₁ = 10r₂
The
heating coils in you toaster during the first five seconds after you turn the
toaster on is Δ K = W + Q.
<span>Your automobile just before you fill
it with gas until you pull away from the gas station at speed v is
ΔK
+ ΔU + ΔEint = W + Q + TMW + TMT</span>
<span>your
body while you sit quietly and eat a peanut butter and jelly sandwich for lunch
is
ΔEint = Q + TET + TER</span>
<span>
your home during five
minutes of a sunny afternoon while the temperature in the home remains fixed is
ΔU = W + Q + TMW + TMT</span>
Answer:
a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm
b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm
Explanation:
Given data:
L = 20 mm
F = 250 N
r₁ = 10 mm
r₂ = 15 mm
v = 0.3
E = 2.07x10⁵ MPa
a) The maximum contact pressure is:
The width of contact is:
b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is
T = 0.3*P = 0.3 * 274.58 = 82.37 MPa
At a distance of
0.8*b = 0.8*0.029 = 0.023 mm
