What is the velocity of an 80 kg person skating across the ice with a kinetic energy of 16, 000 joules?
1 answer:
Answer:
V = 20 m/s
Explanation:
Given the following data;
Mass = 80kg
Kinetic energy = 16,000 joules
To find the velocity;
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
Where;
K.E represents kinetic energy measured in Joules.
M represents mass measured in kilograms.
V represents velocity measured in metres per seconds square.
Substituting into the formula, we have;
16000 = ½*80*V²
16000 = 40V²
V² = 16000/40
V² = 400
Taking the square root of both sides, we have;
V = 20 m/s
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Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.
where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
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'