Answer:
In the table, 1=46.7 °C, 1=165 J, 2=819 J, 3=1510 J, and 4=2830 J.
Other experiments determine that the material has a temperature of fusion of
fusion =235 °C and a temperature of vaporization of vapor=481 °C.
If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l
Answer: A material that does not easily allow a charge to pass through it is called an Plastic and rubber are good insulators. Many types of electric wire are covered with plastic, which insulates well. The plastic allows a charge to be conducted from one end of the wire to the other, but not through the sides of the wire.
Explanation:
Answer:
244mm
Explanation:
I₁ = 3.35A
I₂ = 6.99A
μ₀ = 4π*10^-7
force per unit length (F/L) = 6.03*10⁻⁵N/m
B = (μ₀ I₁ I₂ )/ 2πr .........equation i
B = F / L ..........equation ii
equating equation i & ii,
F / L = (μ₀ I₁ I₂ )/ 2πr
Note F/L = B = F
F = (μ₀ I₁ I₂ ) / 2πr
2πr*F = (μ₀ I₁ I₂ )
r = (μ₀ I₁ I₂ ) / 2πF
r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵
r = 1.4713*10⁻⁵ / 6.03*10⁻⁵
r = 0.244m = 244mm
The distance between the wires is 244m
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:
PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m
Therefore, the displacement would be 0.32 m.