Answer:
6.75m/s
Explanation:
using the second equation of motion, the time is calculated.
and with the formula a= (v - u)/t
where a is acceleration but in this case it's deceleration (and should be negated as you solve it ) .
v is final velocity
u is initial velocity
t is time taken
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:
Now,
Electric field at the mid-point due to charge Q' is given by:
Now,
The net Electric field is given by:
D.) 5kg
This is a trick question. The mass of an object does not change based in location. However the weight of an object does change, this is because Weight = Mass × Gravity. Also mass is measured in kilograms and so the answer is 5 kg. So if you ever want to lose weight just go to the moon!
V0=0m/s (initially at rest)
t=6,7s
s=1/4mi=402,336m
s=(a*t^2)/2 -> a=2*s/t^2 -> a=2*402,336m/(6,7s)^2
a=804,672/44,89=17,93 m/s^2
<span>v=v0+at
</span>v=0+17,93 m/s^2 * 6,7s = 120,131 m/s = 432,4716 km/h
