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
In this case, the movement is uniformly delayed (the final
rapidity is less than the initial rapidity), therefore, the value of the
acceleration will be negative.
1. The following equation is used:
a = (Vf-Vo)/ t
a: acceleration (m/s2)
Vf: final rapidity (m/s)
Vo: initial rapidity (m/s)
t: time (s)
2. Substituting the values in the equation:
a = (5 m/s- 27 m/s)/6.87 s
3. The car's acceleration is:
a= -3.20 m/ s<span>^2</span>
Answer:
In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.
Explanation:
Answer:
v=3.66,h-3.66
Explanation:
vertical = 10sin60 - 10sin 30
horizontal =10cos60 + 10cos 30
v = 10×0.8660-10×0.5
h = 10×0.5 + 10 × 0.8660
v=8.660-5.0 = 3.66
h= 5.0-8.660 = -3.66
The answer should be speed hope this helps
