Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × 
m/s
uniform electric field E = 1.18 × 
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × 
kg ×a = 1.18 × × 1.602 × 
a = 20.75 × 
m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 × (t)
t = 11.80 × 
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.
Answer:
the acceleration is reduced by gravity
a = (15 / .35) - [9.8 * sin(65º)]
Explanation:
break the launch vector into two components, vertical and horizontal
Force Net Vertical=-9.8*.350+15cos65 N
force net horizonal=15sin65
initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i
using i,j vectors..
1 mA = 0.001 A
Therefore, 5 mA = 0.001 * 5
=0.005 A
Resistance = voltage / current
= 100 / 0.005
= 20000 ohms
Current = voltage / resistance
= 25 / 20000
= 0.00125 A (or) 1.25 mA
