Find the velocity of a runner who has a mass of 75 kg and 3,700 Joules of energy.
1 answer:
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Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
<u>a) calculating work done by the force over the journey of the train</u>
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
<u>b) calculate the speed of the train at the end of its journey</u>
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
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