Given:
m(mass of the box)=10 Kg
t(time of impact)=4 sec
u(initial velocity)=0.(as the body is initially at rest).
v(final velocity)=25m/s
Now we know that
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration acting on the body
t is the time of impact
Substituting these values we get
25=0+a x 4
4a=25
a=6.25m/s^2
Now we also know that
F=mxa
F=10 x6.25
F=62.5N
Answer:
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Explanation:
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Answer:
B= 55.6×10^(-7) Tesla
Explanation:
B= μoI/(2πr)
B: magnetic field strength
μo: permeability of free space and is equal to 4π×10^(-7) T.m/A
r: distance from the wire
I : current in the wire
B= (4π×10^(-7)×125)/(2π×4.5)
B= 55.6×10^(-7) Tesla
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