Answer:
32 °C.
Explanation:
Hola.
En este caso, debemos entender que la relación entre el calor y la temperatura viene dada por:

De este modo, dado que estamos estudiando la misma sustancia (agua) con masa constante, la relación calor-temperatura es lineal y directamente proporcional, por tal razón, si se duplica el calor suministrado, la temperatura también será duplicada, de modo que:

¡Saludos!
1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.
Answer:
By the use of slow motion camera.
Explanation:
Visually, it is very hard to differentiate between an ac and dc power supply. But Since, we that In Ac supply polarity changes 100 times in a second ( because frequency of ac supply is 50 Hz generally). Whereas, Dc gives a steady power supply. So, in slow motion camera we can easily capture the flickering light tubes which won't happen in case of dc supply.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is
= r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius
and mass
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)

Hence the ratio is 
[ ans]
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr