So the problem are asking to find the value of G base on the formula of the said equation of the magnitude of gravitational attraction on either body. Base on that, the possible answer or the derived formula of the said function is G = Fr^2/m1m2. I hope you are satisfied with my answer and feel free to ask for more
Initial speed = 56mph
Final speed = 35mph
Time taken = 6.7seconds...
Converting the time to hour.. Divide by 3600..
= 6.7/3600
=0.00186hour..
Acceleration = v-u/t
a = 35-56/0.00186
a = -11283.6mph²
The negative sign shows that it decelerated...
V² = u²+2as
(35)² = (56)² + 2×-11283.6×s
Where s is the distance covered within that time...
1225 = 3136 - 22567.2s
22567.2s = 3136-1225
22567.2s = 1911
S = 1911/22567.2
S = 0.08468miles...
But at the end of the question we were made to understand that 1miles = 5280ft
Therefore 0.08468miles = (0.08468×5280)ft
= 447. 11feets...
Which is approximately 447ft.....
Hope this helped.... ?
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
An increase in voltage causes the flow of electric current to INCREASE. An increase in resistance causes the flow of electric current to DECREASE.
100% right
Answer:
Explanation:
Answer:
Explanation:
Given that,
System of two particle
Ball A has mass
Ma = m
Ball A is moving to the right (positive x axis) with velocity of
Va = 2v •i
Ball B has a mass
Mb = 3m
Ball B is moving to left (negative x axis) with a velocity of
Vb = -v •i
Velocity of centre of mass Vcm?
Velocity of centre of mass can be calculated using
Vcm = 1/M ΣMi•Vi
Where M is sum of mass
M = M1 + M2 + M3 +...
Therefore,
Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb
Rearranging for better understanding
Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)
Vcm = (m•2v + 3m•-v) / (m + 3m)
Vcm = (2mv — 3mv) / 4m
Vcm = —mv / 4m
Vcm = —v / 4
Vcm = —¼V •i