Resistance = ρ * (L/A) and Rf = Ri * ([1 + α * (Tf – Ti)]
ρ = Resistivity L = length in meters A = cross sectional area in m^2 α = temperature coefficient of resistivity
L = 1.50 m Area = π * r^2 r = d/2 = 0.25 cm = 2.5 * 10^-3 m Area = π * (2.5 * 10^-3)^2
The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases. Resistance = Voltage ÷ Current At 20˚, R = 15 ÷ 18.5 At 92˚, R = 15 ÷ 17.2
Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures. Resistance = ρ * (L/A) ρ = Resistance * (A/L)
At 20˚, ρ = (15 ÷ 18.5) * [π * (2.5 * 10^-3)^2] ÷ 1.5 = At 92˚, ρ = (15 ÷ 17.2) * [π * (2.5 * 10^-3)^2] ÷ 1.5 =
Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod.
Rf = Ri * ([1 + α * (Tf – Ti)] Rf = 15 ÷ 17.2, Ri = 15 ÷ 18.5, Tf = 92˚, Ti = 20˚
15 ÷ 17.2 = 15 ÷ 18.5 * [1 + α * (92 – 20)] Multiply both sides by (18.5 ÷ 15) (18.5 ÷ 15) * (15 ÷ 17.2) = 1 + α * 72 Subtract 1 from both sides (18.5 ÷ 15) * (15 ÷ 17.2) – 1 = α * 72 Divide both sides by 72 α = 1.05 * 10^-3
Kinetic friction, for example, generally turns energy into heat, and although we associate kinetic friction with energy loss, it really is just a way of transforming kinetic energy into thermal energy. The law of conservation of energy applies always, everywhere, in any situation. ?
Answer:
The potential energy of the ball is 784 joules. And the kinetic energy of it is 392 while falling halfway down.
Explanation:
PE = mass (2kg) * Gravitational acceleration (9.8 m/s^2)* height (40 meters)
KE = 1/2 mass (1 kg) * velocity^2 (19.8)
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Answer:
A force has both magnitude and direction, therefore: Force is a vector quantity; its units are newtons, N. Forces can cause motion; alternatively forces can act to keep (an) object(s) at rest. ... Consider two forces of magnitudes 5 N and 7 N acting on a particle, with an angle of 90◦ between them.
Explanation:
from google
