C
It does not affect the brightess of a lightbulb
Complete Question
One arm of a U-shaped tube (open at both ends) contains water, and the other alcohol.
If the two fluids meet at exactly the bottom of the U, and the alcohol is at a height of 18 cm, at what height will the water be?Assume the density of alcohol is ![\rho_a = 790\ kg/m^3](https://tex.z-dn.net/?f=%5Crho_a%20%20%3D%20%20790%5C%20%20kg%2Fm%5E3)
Answer:
The height of water is
Explanation:
From the question we are told that
The height of the alcohol is ![h_a =18 \ cm = 0.18 \ m](https://tex.z-dn.net/?f=h_a%20%20%3D18%20%5C%20cm%20%20%3D%20%200.18%20%5C%20m)
The density of the alcohol is ![\rho_a = 790\ kg/m^3](https://tex.z-dn.net/?f=%5Crho_a%20%20%3D%20%20790%5C%20%20kg%2Fm%5E3)
Generally the pressure on both arm of the tube are equal given that they are both open
i,e ![P_a = P_w](https://tex.z-dn.net/?f=P_a%20%20%3D%20%20P_w)
Where
is pressure of alcohol and
is pressure of water
So the pressure on the arm of the tube containing the alcohol is mathematically evaluated as
![P_a = g * h * \rho](https://tex.z-dn.net/?f=P_a%20%20%3D%20%20g%20%20%2A%20%20h%20%20%2A%20%20%5Crho)
substituting values
![P_a =9.8 * 0.18 * 790](https://tex.z-dn.net/?f=P_a%20%20%3D9.8%20%20%2A%20%200.18%20%2A%20%20790)
![P_a = 1394 \ Pa](https://tex.z-dn.net/?f=P_a%20%20%3D%201394%20%5C%20Pa)
Generally the pressure on the arm of the tube containing the water is mathematically evaluated as
![P_w = g * h_w * \rho_w](https://tex.z-dn.net/?f=P_w%20%20%3D%20%20g%20%2A%20%20%20h_w%20%20%20%2A%20%20%5Crho_w)
where
is the density of water which has a value ![\rho _w = 1000 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20_w%20%20%3D%20%201000%20%5C%20kg%2Fm%5E3)
So
![1394 = 9.8 * h_w * 1000](https://tex.z-dn.net/?f=1394%20%3D%209.8%20%20%2A%20%20h_w%20%20%2A%20%201000)
=> ![h_w = \frac{1394}{9800}](https://tex.z-dn.net/?f=h_w%20%20%3D%20%20%5Cfrac%7B1394%7D%7B9800%7D)
=> ![h_w = 0.142 \ m](https://tex.z-dn.net/?f=h_w%20%20%3D%20%200.142%20%5C%20m)
D. Substances that atoms make up.
Answer:
The angle of the launch is 17.09 degrees.
Explanation:
Given that,
The initial speed of a cannon is 100 m/s
It is launched at some angle and it lands after being in the air for 6 s.
We need to find the angle of the launch.
The time taken by the projectile to reach the ground is called the time of flight. It is given by the formula as follows :
![T=\dfrac{2u\sin\theta}{g}](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7B2u%5Csin%5Ctheta%7D%7Bg%7D)
Here,
= launch angle
![\sin\theta=\dfrac{Tg}{2u}\\\\\sin\theta=\dfrac{6\times 9.8}{2\times 100}\\\\\theta=\sin^{-1}(0.294)\\\\\theta=17.09^{\circ}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7BTg%7D%7B2u%7D%5C%5C%5C%5C%5Csin%5Ctheta%3D%5Cdfrac%7B6%5Ctimes%209.8%7D%7B2%5Ctimes%20100%7D%5C%5C%5C%5C%5Ctheta%3D%5Csin%5E%7B-1%7D%280.294%29%5C%5C%5C%5C%5Ctheta%3D17.09%5E%7B%5Ccirc%7D)
Hence, the angle of the launch is 17.09 degrees.
In physics, the amount of energy put out or produced in a given amount of time. Power<span> is often measured in watts or kilowatts. In mathematics, a</span>power<span> is a number multiplied by itself the number of times signified by an exponent placed to the right and above it.</span>