“Don't hand that holier than thou line to me” is what the asymptote
said to the removable discontinuity.
The distance between the
curve and the line where it approaches zero as they tend to infinity is the line in the asymptote
of a curve. This is unusual for modern authors but in some
sources the requirement that the curve may not cross the line infinitely often
is included.
The point that does not fit the rest of the graph or is
undefined is called a removable discontinuity. By filling in a single
point, the removable discontinuity can be made connected.
Answer:
14,700 N
Explanation:
The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:
(1)
the weight of the hyppo is

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

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:
0.686 g of ice melts each second.
Solution:
As per the question:
Cross-sectional Area of the Copper Rod, A = 
Length of the rod, L = 19.6 cm = 0.196 m
Thermal conductivity of Copper, K = 
Conduction of heat from the rod per second is given by:

where
= temperature difference between the two ends of the rod.
Thus

Now,
To calculate the mass, M of the ice melted per sec:

where
= Latent heat of fusion of water = 333 kJ/kg

The mass of water that must be raised is 
Explanation:
Since the process is 70% efficiency, the power in output to the turbine can be written as

where
is the power in input.
The power in input can be written as

where
W is the work done in lifting the water
t = 3 h = 10,800 s is the time elapsed
The work done in lifting the water is given by

where
m is the mass of water
is the acceleration of gravity
h = 45 m is the height at which the water is lifted
Combining the three equations together, we get:

Where

And solving for m, we find:

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