Answer:
The variation rate is 5.42 10⁻⁵ cm²/ºC
Explanation:
When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of a rectangle
a = L W
They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature
da / dT = d(LW) / dt
We use the derivative of a product since the two magnitudes change
da / dT = W dL/dT + L dW/dT
The values they give us are
= 1.9 10⁻⁵ cm/ºC
= 8.5 10⁻⁶ cm/ºC
W = 1.6 cm
L= 2.8 cm
Substituting the values and calculating
= 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶
= 3.04 10⁻⁵ + 2.38 10⁻⁵
= 5.42 10⁻⁵ cm²/ºC
The variation rate is 5.42 10⁻⁵ cm²/ºC
Answer : The average speed of the sprinter is, 34.95 Km/hr
Solution :
Average velocity : It is defined as the distance traveled by the time taken.
Formula used for average velocity :
where,
= average velocity
d = distance traveled = 200 m
t = time taken = 20.6 s
Now put all the given values in the above formula, we get the average velocity of the sprinter.
conversion :
(1 Km = 1000m)
(1 hr = 3600 s)
Therefore, the average speed of the sprinter is, 34.95 Km/hr
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
527 Hz
Solution:
As per the question:
Beat frequency of the player,
Frequency of the tuning fork, f = 523 Hz
Now,
The initial frequency can be calculated as:
when
when
But we know that as the length of the flute increases the frequency decreases
Hence, the initial frequency must be 527 Hz