Where is the highest-energy electron found in an atom of hydrogen?
1 answer:
You might be interested in
This question is incomplete, the complete question is;
The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters
a) Find the volume V in terms of x and L
b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.
Answer:
a) the volume V in terms of x and L is ((√3/4)x²L) m³
b) required expression is (2/(3)^(1/u))√(At/L)
Explanation:
Given that;
form the question and image below;
triangular prism ends are equilateral triangle
side length = x meter
Dimension of the prism = L meter
Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter
Volume of the triangular prism = Area × height
= √3/4 (x)² × L
V = ((√3/4)x²L) m³
Therefore, the volume V in terms of x and L is ((√3/4)x²L) m³
b)
Rate of water entering = A m³/hr
Depth of water tank = d meter
Time = t
Length of prism = L
now Rate of water entering is A m³/hr
dv/d = A [ V = ((√3/4)x²L) m³ ]
and
dv/dt = √3/4 [2x dx/dt ] L { L is constant }
so
A = √3/4 [2x dx/dt ] L
∫A dt = √3/2 [ Lx dx ] { Integrate both sides}
At = √3/2 × Lx × x²/2
x² = uAt / √3L { we find square root of both sides}
x = √( uAt / √3L )
x = (2/(3)^(1/u))√(At/L)
Therefore; required expression is (2/(3)^(1/u))√(At/L)
Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q =
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
