Answer: 1.3617×10^-14 N
Explanation: The formulae that relates electric field intensity, magnitude of charge and force is given by the formulae below.
F = Eq
Where F = electric force =?
E = strength of electric field = 88000 N/c
q = magnitude of proton charge = 1.602×10^-19 c
By substituting the parameters, we have that
F = 88000 × 1.602×10^-19
F = 13.617×10^-15
F = 1.3617×10^-14 N
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
WE CALCULATE POWER AND RATE OF DOING WORK IS CALLED POWER
Answer:
2.5 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s²
For circular motion, the expression for acceleration is given as,
a = ω²r ................ Equation 1
Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.
Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.
Substitute into equation 1
a = 5²(0.10)
a = 25(0.10)
a = 2.5 m/s²
Hence the acceleration of the particle = 2.5 m/s²
Hence, none of the option is correct
(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second
<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
