Answer:
1. 24375 N/C
2. 2925 V
Explanation:
d = 12 cm = 0.12 m
F = 3.9 x 10^-15 N
q = 1.6 x 10^-19 C
1. The relation between the electric field and the charge is given by
F = q E
So,
E = 24375 N/C
2. The potential difference and the electric field is related by the given relation.
V = E x d
where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.
By substituting the values, we get
V = 24375 x 0.12 = 2925 Volt
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Spring C stretches 100 cm.
Explanation:
The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.
Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.
According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"
F = ke
k is the spring constant
e is the extension
We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.
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Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
<h3>
Answer:</h3>
117.6 Joules
<h3>
Explanation:</h3>
<u>We are given;</u>
- Force of the dog is 24 N
- Distance upward is 4.9 m
We are required to calculate the work done
- Work done is the product of force and distance
- That is; Work done = Force × distance
- It is measured in Joules.
In this case;
Force applied is equivalent to the weight of the dog.
Work done = 24 N × 4.9 m
= 117.6 Joules
Hence, the work done in lifting the dog is 117.6 Joules