The correct answer to the question is : D) 352.6 m/s.
CALCULATION :
As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.
We are asked to calculate the velocity of sound at 36 degree celsius.
Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.
The velocity at this temperature is calculated as -
V = 331 + 0.6T m/s
= 331 + 0.6 × 36 m/s
= 331 + 21.6 m/s
= 352.6 m/s.
Here, T denotes the temperature of the surrounding.
Hence, velocity of the sound will be 352.6 m/s.
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.
The moon has a small amount of gravity. Low tides mean the moon is not pulling on the water. High tides mean that the moon is pulling on the water.
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''


Vy in this problem will follow this equation =

where g is the gravity acceleration

This is equation (1)
For Y(t) :

We suppose yi = 0

This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s

The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m