Answer:
0.15
Explanation:
Assuming the rope is horizontal, sum the forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum the forces in the x direction:
∑F = ma
F − Nμ = ma
Substitute:
F − mgμ = ma
mgμ = F − ma
μ = (F − ma) / (mg)
Plug in values:
μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)
μ = 0.15
Its b i literally have had this exact question
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns


Therefore, the number of turns of wire needed is 573.8 turns
To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by

where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as


Our values are given as,
Total mass is
= 200lb ,however the mass of solid vegetable and water is given as,



Replacing at our equation we have,



Therefore the heat removed is 22411.2 Btu