As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m
Consider the cross section to the right with an overall height h 4.2 inch, bottom flange plate width b1-2.5 inch, top flange pla
te width b2-2.0 inch, and the web and both flange plates have a thickness of t-0.25 inch. The center of coordinate system is located at the centroid of the cross-section area. a) Find the y distance of centroid for the cross section to the bottom edge in inch. b) Find the Sx section modulus in inch3. c) The section is rotated counter-clockwise 20 degrees about its centroid. Find the rotated l moment of inertia in inch4. d) Find the rotated Iky product of inertia in inch4.
1 answer:
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To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
Therefore the velocity right after catching the ball is 0.0975m/s
Answer:
The energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
So, the energy in its ground state is 10 meV. Hence, this is the required solution.