Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.
Answer:
+ 3.0 m
Explanation:
displacement is shortest distance from fixed point O in particular direction . in diagram shortest distance at end from O is 3 m and it is right of O so +. HENCE +3.0m
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:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
