Average speed = distance traveled / time
average speed = (126.5 m * 3.5 laps) / (4.17 min)
= 106.2 m/min
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer:
a) B = 1.99 x 10⁻⁴ Tesla
b) B = 0.88 x 10⁻⁴ Tesla
Explanation:
According to Biot - Savart Law, the magnetic field due to a currnt carrying straight wire is given as:
B = μ₀ I L/4πr²
where,
μ₀ = permebility of free space = 1.25 x 10⁻⁶ H m⁻¹
I = current = 2 A
L = Length of wire = 40 cm = 0.4 m
a)
r = radius of magnetic field = 2 cm = 0.02 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.02 m)²
<u>B = 1.99 x 10⁻⁴ Tesla</u>
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b)
r = radius of magnetic field = 3 cm = 0.03 m
Therefore,
B = (1.25 x 10⁻⁶ H m⁻¹)(2 A)(0.4 m)/4π(0.03 m)²
<u>B = 0.88 x 10⁻⁴ Tesla</u>
