Answer:
1.3636
Explanation:
Write the expression for the refractive index.
n=c/v
c= speed of light in air
v= speed of light in material
=(3×10^8 m/s)/(2.2×10^8 m/s)
=1.3636
Normally, the number of electrons is equal to the number of protons, which makes atoms electrically neutral. The number of protons in an atom is the defining feature of an atom. It's what makes one element different from another
HOPE THIS HELPS!!!
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres
