Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
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
Answer:
(a) 29 cm
(b) 43.5 cm
Explanation:
(a) when loop A is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 70 cm mark
-2 N at x
Taking the sum of the torques about B:
∑τ = Iα
(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0
18 Ncm − 2 N (x − 70 cm) = 0
2 N (x − 70 cm) = 18 Ncm
x − 70 cm = 9 cm
x = 79 cm
The distance from the center is |50 cm − 79 cm| = 29 cm.
(b) when loop B is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 20 cm mark
-2 N at x
Taking the sum of the torques about A:
∑τ = Iα
(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0
-27 Ncm − 2 N (x − 20 cm) = 0
2 N (x − 20 cm) = -27 Ncm
x − 20 cm = -13.5 cm
x = 6.5 cm
The distance from the center is |50 cm − 6.5 cm| = 43.5 cm
Answer:
1) a rubber band
2) the spring of retractable pen
3) a spring loaded toy gun
Explanation:
Hooke's law states that; provided the elastic limit of a material is not exceeded, the force exerted on an elastic material is directly proportional to its extension. This relationship was first captured by Robert Hooke in 1660 when he asserted that 'as the extension, so is the force!'.
Hooke's law generally deals with elastic or stretchable materials. These materials can be deformed, but returned to their original shapes when the deforming force is removed. This deforming force causes an extension in the material which is directly proportional to the deforming force. That is F= Kx where K is the called the force constant, F is the deforming force and x is the magnitude of extension brought about by the force.
Various real life applications of Hooke's law have been listed in the answer. Any material that makes use of a loaded spiral spring or indeed any kind of elastic material obeys Hooke's law.
They think everything is involved with tech , some business like going old school , it’s artificiaciality is off meaning consequences aren’t actually accurate like in real life , lack of flexibility can’t change anything because it’s programmed so you can’t actually act question or change scenarios ,it costs too much as well , can be health risk can cause stress and anxiety,
