Answer: <span>D. A bimetallic strip bends so that the steel is on the outside curve
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When something has an increased temperature, its volume will expand. Then, if the temperature drops, its volume should be smaller. From there option A and B are out since the liquid in thermometer is expand or move up.
When you put two kinds of different metal with a different coefficient of thermal expansion, the outer curve metal will be the one with lesser coefficient when temperature drop. Since the question about drop in temperature then the metal should be bend
Brass will expand 1.5 times more than the steel so the outer curve would be the steel.
No, the object's displacement and distance travelled will be equal, but since the initial position is unknown, the object's position might not match up with its displacement and distance travelled.
We cannot assert that the displacement or distance equals the position because the initial position is not provided. We could reach a different conclusion if the starting position had been zero because the distance from zero is equal to the position.
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We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

(parallel direction) (1)

(perpendicular direction) (2)
where

is the angle of the inclined plane, N is the normal reaction of the plane,

is the frictional force, with

being the coefficient of friction.
From eq.(2), we find

and if we substitute into eq.(1), we can find the acceleration of the block:

from which
Answer:
a) 0.15 μC b) 9.4*10¹¹ electrons.
Explanation:
As the total charge must be conserved, the total charge on the spheres, after being brought to contact each other, and then separated, must be equal to the total charge present in the spheres prior to be put in contact:
Q = +8.2μC +9.0 μC +(-7.8 μC) + (-8.8 μC) = +0.6 μC
As the spheres are assumed perfect conductors, as they are identical, once in contact each other, the excess charge spreads evenly on each sphere, so the final charge, on each of them, is just the fourth part of the total charge:
Qs = Qt/4 = 0.6 μC / 4 = 0.15 μC.
b) As the charge has a positive sign, this means that each sphere has a defect of electrons.
In order to know how many electrons are absent in each sphere, we can divide the total charge by the charge of one electron, which is the elementary charge e, as follows:
