Answer:

Explanation:
It is given that,
Mass of Albertine, m = 60 kg
It can be assumed, the spring constant of the spring, k = 95 N/m
Compression in the spring, x = 5 m
A glass sits 19.8 m from her outstretched foot, h = 19.8 m
When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,




So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Answer:
The correct answer is Dean has a period greater than San
Explanation:
Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.
T² = (4π² / G M) r³
When applying this equation to our case, the planet with a greater orbit must have a greater period.
Consequently Dean must have a period greater than San which has the smallest orbit
The correct answer is Dean has a period greater than San
Answer:
Q = - 4312 W = - 4.312 KW
Explanation:
The rate of heat of the concrete slab can be calculated through Fourier's Law of heat conduction. The formula of the Fourier's Law of heat conduction is as follows:
Q = - kA dt/dx
Integrating from one side of the slab to other along the thickness dimension, we get:
Q = - kA(T₂ - T₁)/L
Q = kA(T₁ - T₂)/t
where,
Q = Rate of Heat Loss = ?
k = thermal conductivity = 1.4 W/m.k
A = Surface Area = (11 m)(8 m) = 88 m²
T₁ = Temperature of Bottom Surface = 10°C
T₂ = Temperature of Top Surface = 17° C
t = Thickness of Slab = 0.2 m
Therefore,
Q = (1.4 W/m.k)(88 m²)(10°C - 17°C)/0.2 m
<u>Q = - 4312 W = - 4.312 KW</u>
<u>Here, negative sign shows the loss of heat.</u>