Object A and object B are initially uncharged and are separated by a distance of 1 meter. Suppose 10,000 electrons are removed f
1 answer:
Answer:
1
Explanation:
Let’s denote the charge magnitude for 10,000 electrons as .
When object A loses charge q, it results in a total charge of (since electrons are negatively charged, their loss turns A positively charged). On the other hand, B, which gains q, assumes a charge of
. They are initially 1 meter apart.
According to Coulomb’s law, the force acting between them is
When an additional 10,000 electrons are stripped from A, its charge becomes , while B’s charge becomes
. Now, they are 2 meters apart.
The force at this new distance is
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Answer:
a) Blood mass is
b) The count of blood cells is
Explanation:
From the problem statement, we learn that
The blood volume is
The density of blood is
% of blood which consists of cells is = 45.0%
the % of blood that is plasma is = 55.0%
density of blood cells is
% of cells that are white is = 1%
% of cells that are red is = 99%
The red blood cell diameter is =
The red blood cell radius is
The mass is generally represented mathematically as
Substituting values
Cell mass is = 45% of m
The white blood cells volume is = 1% of the cells volume
The volume of a single cell is
The red blood cells volume is
The total red blood cell count is
The total white blood cell count is
The overall number of blood cells is
Answer:
(a) the coefficient of friction is 0.451
This was derived using the energy conservation principle (the total energy in a closed system remains constant).
(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.
Explanation:
For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.
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Answer:
0.0984
Explanation:
The first diagram below illustrates a free body diagram that will aid in resolving this problem.
According to the diagram, the force's horizontal component can be expressed as:
Substituting 42° for θ and 87.0° for
Meanwhile, the vertical component is:
Again substituting 42° for θ and 87.0° for
In resolving the vector, let A denote the components in mutually perpendicular directions.
The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ
The frictional force can be expressed as:
Where;
is the coefficient of friction
N = the normal force
Also, the normal reaction (N) is calculated as mg - F sin θ
Substituting . Normal reaction becomes:
By balancing the forces, the horizontal component of the force equals the frictional force.
The horizontal component is described as follows:
Rearranging the equation above to isolate leads to:
Substituting in the following values:
m = 73 kg
g = 9.8 m/s²
Thus:
Therefore, the coefficient of friction is = 0.0984
