A solid metal object hangs from a force sensor by a thread in the open air, and the actual weight is measured to be 0.71 N. Then
1 answer:
Answer:
The apparent weight of the object is 0.465 N.
Explanation:
Given that,
Weight = 0.71 N
Water level = 50 mL
object inserted = 75 mL
We need to calculate the volume of solid
Using formula of volume
We need to calculate the buoyancy force
Using formula of buoyancy force
Put the value into the formula
We need to calculate the apparent weight of the object
Using formula of apparent weight
Put the value into the formula
Hence, The apparent weight of the object is 0.465 N.
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Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
Answer:
The volume of water evaporated is 199mL
Explanation:
Concentration is calculated with the following formula
where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.
So we isolate the variable n to know the amount of moles, using the volume given in liters
Now, we isolate the variable V to know the new volume with the new concentration given.
Finally, the volume of water evaporated is the difference between initial and final volume.