Answer:
La escala del termómetro ''A'' es grados Celsius.
La escala del termómetro ''B'' es grados Fahrenheit.
Explanation:
Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.
Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.
Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :
(I)
Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos 
en la ecuación (I) deberíamos obtener 
⇒
Efectivamente obtenemos el valor esperado. Finalmente, corroboramos que la temperatura del termómetro ''A'' está medida en grados Celsius y la temperatura del termómetro ''B'' en grados Fahrenheit.
Answer:
distance
Explanation:
it is the distance traveled by light in one year
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
Answer:
in the parallel connection the light bulbs shine less than in the series connection
Explanation:
In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is
P = i² R
where R is the resistance of each bulb and i the current of the circuit.
If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,
i = i₁ + i₂
if the two light bulbs are the same
i = 2 i₁
i₁ = i / 2
so the power of each bulb is is
P = i₁² R
P = R i² / 4
P = ¼ P_initial
Therefore we see that in the parallel connection the light bulbs shine less than in the series connection
