The region where warm and cold air masses meet
2 answers:
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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:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N
Therefore, for the weight of the cadet on the exoplanet, W₁, we have;
The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b. so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx = and filling in the time the object was at 26 m/s:
Δx = 0t + so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.