Answer:
The initial velocity of the water from the tank is 5.42 m/s
Explanation:
By applying Bernoulli equation between point 1 and 2
At the point 1
P₁=0 ( Gauge pressure)
V₁= 0 m/s
Z₁=3 m
At point 2
P₂=0 ( Gauge pressure)
Z₂= 0 m/s
Now by putting the values
V₂= 5.42 m/s
The initial velocity of the water from the tank is 5.42 m/s
Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:
In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:
So we can rewrite the energy in the ground state as:
Finally
As we know that range of the projectile motion is given by
here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be
so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree