A car, traveling at , encounters a dip in the road. The radius of curvature at the bottom of the dip is . Each of the car’s four
1 answer:
Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m
Where Fe is elastic force, W the weight and the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
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Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
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where
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f =
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Now, by using lens maker formula:
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Now,
Solving the above eqn, we get:
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Answer:
Explanation:
1) for a given n value the l value can be from 0 to n-1
So if n= 5 it can take 0,1,2,3,4
i.e it can take 5 values
2)for an electron with l =3
it can be from -3 -2 -1 0 1 2 3
i.e it can take 7 values
3) n = 3 !!
l = 0 , 1 , 2
for l=0 , m = 0 total = 1
for l= 1 ,m = -1,0,1 total = 3
for l = 2, m=-2,-1,0,1,2 total = 5
5+3+1 = 9
total possible states = 9 * 2 = 18
Answer is 168
4)given l=3 and n=3
orbital quantum number cannot be equal to principal quantum number
its max value is l-1 only
5)L = sqrt(l(l+1))x h'
for it to be max l should be max
for n = 3 max l value is 2
therfore it is sqrt(2(2+1)) x h'
this is the answer