7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
<u>Disaster</u>:- a calamitous event, especially one occurring suddenly and causing great loss of life, damage, or hardship, as a flood / airplane crash.
:))
Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
now,
inserting the values
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
