If you could repeat the lab and make it better, what would you do differently and why? There are always ways that labs can be im
2 answers:
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As we know that spring force is given as
here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have
so the elastic constant of the spring will be 200 N/m
Work done is 0.442J
<u>Explanation:</u>
Given:
Spring constant, k = 33 N/m
Distance, x₁ = 0.15m
Additional distance, x₂ = 0.072 m
Total distance = 0.15 + 0.072 m
= 0.222 m
Work done, W = ?
We can calculate work done by the formula
On substituting the value we get:
Therefore, work done is 0.442J
Answers:
1) flow of traffic =
2) average speed =
3)density of traffic =
Explanation:
1) to find flow of traffic
we use the relation
where q is the traffic flow and h is average time headway.
insert the value
2) to find average traffic speed
use the relation
where u is average speed S is average spacing and h average time
so inserting the values
3) density of traffic
where K is density of traffic q is flow of traffic and u is average speed
inserting values from above solved parts
Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
Explanation:
Given: Density = 80 (1 n =
m) =
= 1.0 mm (1 mm = 0.001 m) = 0.001 m
= 3.0 mm = 0.003 m
r = 2.0 mm = 0.002 m (from the symmetry axis)
The charge per unit length of the cylinder is calculated as follows.
Substitute the values into above formula as follows.
Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.