Answer:
The force is 
Explanation:
The diagram for this question is shown on the first uploaded image
At Equilibrium the summation of the of force on the vertical axis is zero
i.e 
=> 
is the is the speed of water at the nozzle which can be mathematically evaluated as
substituting 
for R and 
for 
is the is the speed of water at the pipe which can be mathematically evaluated as
substituting for R and 
for 
is he density of water with value 
Substituting values into the equation above
At Equilibrium the summation of the of force on the horizontal axis is zero
i.e 
=> 
Since The speed at both A and B nozzle are the same then remains the same
Substituting values
=> 
Hence the force acting on the flange bolts required to hold the nozzle in place is
Answer:
a)
two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)
b)
both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Explanation:
Given that;
L = 0.26 m
k = 180 N/m
x = 0.039 m
a)
we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.
b)
Spring force F = kx
F = 180 × 0.039
F = 7.02 N
Now, Electrostatic force F = Keq²/r²
where r = L + x = ( 0.26 + 0.039 )
we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2
so from the equation; F = Keq²/r²
Fr² = Keq²
q = √ ( Fr² / Ke )
we substitute
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )
q = √ (0.627595 / 9×10⁹)
q = √(6.97 × 10⁻¹¹)
q = 8.35 × 10⁻⁶ C
Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C
Answer:
A bicycle on the top of the hill has the highest potential energy, and when the bike goes down, it transfers to kinetic because it is moving
Explanation:
yeah
Answer:
Explanation:
In case of diffraction , angular width of central maxima =2 λ/d
λ is wave length of light and d is slit width
In case of interference , angular width of each fringe
= λ /D
D is distance between two slits
No of interference fringe in central diffraction fringe
=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.
