Answer:
4 Four objects are stiuated along the y axis as follow: a 20 kg object is at +3 m, a 3 kg object is at +2.5 m, a 2.5 kg object i
1 answer:
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Answer:
Explanation:
First of all, I used the specific heat of water as 4182 J/(kgC) and the specific heat of ethyl alcohol (EtOH) as 2440 J/(kgC); that means that we need the masses in kg, not g.
120.g = .1200 kg of ethyl alcohol. Now for the formula:
where spheat is specific heat.
Filling that horrifying-looking formula in with some values:
and
and
16(4182x + 292.8) = 83640x + 2928 and
66912x + 4684.8 = 83640x + 2928 and
1756.8 = 16728x so
x = .105 kg and the amount of water added is 105 g
Answer:
(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons
(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters
Explanation:
The given parameters are;
The mass of particle, A, m₁ = 2 kg
The mass of particle, B, m₂ = 0.3 kg
The mass of particle, C, m₃ = 0.05 kg
The distance between particle 'A' and particle 'B', r₁ = 0.15 m
The distance between particle 'B' and particle 'C', r₂ = 0.05 m
(a) The gravitational force, 'F', is given as follows;
Where;
F = The force between the two masses
G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²
m₁ = The mass of object 1
m₂ = The mass of object 2
If 'C' is placed at 0.05 m from 'B', we have;
F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰
The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)
F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰
The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)
The force, 'F', acting on object 'C' = F₁₃ - F₂₃
F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N
The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N
(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x
We have;
F₂₃ = F₁₂
By plugging in the values and removing like terms, we get;
(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²
0.1·x² - 0.23·x + 1.3225 = 0.015·x²
0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0
0.085·x² - 0.23·x + 0.13225= 0
x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))
x ≈ 0.829, or x ≈ 1.877
Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'
Acceleration of the table: B. 0.50 meters/second2
Explanation:
The problem can be solved by using Newton's second law of motion, which states that the net force acting on an object is the product of its mass and its acceleration. Mathematically:
where
is the net force
m is the mass
a is the acceleration
For the table in this problem, we have:
is the net force on the table, because there are two forces of 125 N and 120 N acting in opposite directions
m = 10.0 kg is the mass of the table
Solving for a, we find the acceleration:
Learn more about Newton's second law:
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