The closer to the equator, the hotter the climate will be.
Answer:
600Hz
Explanation:
In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.
In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz
Answer:
The change in temperature is
Explanation:
From the question we are told that
The temperature coefficient is 
The resistance of the filament is mathematically represented as
![R = R_o [1 + \alpha \Delta T]](https://tex.z-dn.net/?f=R%20%20%3D%20%20R_o%20%5B1%20%2B%20%5Calpha%20%20%5CDelta%20T%5D)
Where
is the initial resistance
Making the change in temperature the subject of the formula
![\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BR%7D%7BR_o%7D%20-%201%20%5D)
Now from ohm law

This implies that current varies inversely with current so

Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7BI%7D%20-%201%20%5D)
From the question we are told that

Substituting this we have
![\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]](https://tex.z-dn.net/?f=%5CDelta%20T%20%20%3D%20%5Cfrac%7B1%7D%7B%5Calpha%20%7D%20%5B%5Cfrac%7BI_o%7D%7B%5Cfrac%7BI_o%7D%7B8%7D%20%7D%20-%201%20%5D)
=> 
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
The 2 means there are two phosphate ions in a molecule of magnesium phosphate.