To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,






Therefore the percent of the atom's volume is occupied by mass is 
Answer:
In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.
Explanation:
The meaning of odd nuclei is atomic mass is odd.
A=odd number.
A=Z+n
Here, Z is proton either it will odd or n will odd which is neutron.
Now according to the shell model the left out proton or neutron will contribute to the spin and parity.
For example,
Take the case of isotope of nitrogen-15.
Here Z is 7, and n is 8 will not contribute in spin.
Now, for Z=7.

Here,

and, L=1.
Fort parity,

Put the value of L.
Parity will be -1.
Now, spin will be
.
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Part g)

Explanation:
Initial speed of the launch is given as
initial speed = 
angle =
degree
Now the two components of the velocity

similarly we have

Part a)
Now we know that horizontal range is given as

maximum height is given as

so we have

time of flight is given as



Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as



Part c)
Initial vertical velocity is given as


Part d)
Initial speed is given as

so we will have


Part e)
Angle of projection is given as



Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as


Part g)
For maximum range the angle should be 45 degree
so maximum range is


Answer:
E) 800 km/h
Explanation:
The computation of the average vector velocity module of the plane, at that time is shown below:
The displacement vector is
d^2 = d1^2 + d2^2
where,
d1 = northeast displacement
d2 = southeast displacement
Now
d^2 = (120)^2 + (160)^2
= 14400 + 2560
= 40000
= √40000
d = 200 km
Now the average velocity is
V = ΔS ÷ Δt
= 200 ÷ 1 ÷ 4
= 200 × 4
= 800 km/h
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s