Answer:
Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]
Explanation:
To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.
The gravitational force is equal to:
Fg = (10 * 9.81) = 98.1 [N]
Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.
N - Fg = 0
N = Fg
N = 98.1 [N]
The friction force is defined as the product of normal force by the coefficient of friction.
Ff = N * μ
Ff = 98.1 * 0.4
Ff = 39.24 [N]
By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.
60 - Ff = m * a
60 - 39.24 = 10 * a
a = 2.076[m/^2]
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure difference of the first bubble is ![\Delta P _1 =10 J/m^3](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_1%20%3D10%20%20J%2Fm%5E3)
The pressure difference of the second bubble is ![\Delta P _2 =20 J/m^3](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_2%20%3D20%20%20J%2Fm%5E3)
The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger
Explanation:
From the question we are told that
The radius of the first bubble is ![r_1 = 10 \ mm=0.01 \ m](https://tex.z-dn.net/?f=r_1%20%3D%20%2010%20%5C%20mm%3D0.01%20%5C%20m)
The radius of the second bubble is ![r_2 = 5 \ mm = 0.005 \ m](https://tex.z-dn.net/?f=r_2%20%20%3D%20%205%20%5C%20mm%20%20%3D%20%200.005%20%5C%20m)
The surface tension of the soap solution is ![s = 25 \ mJ/m^2 = 25*10^{-3} J/m^2](https://tex.z-dn.net/?f=s%20%3D%20%2025%20%5C%20mJ%2Fm%5E2%20%3D%2025%2A10%5E%7B-3%7D%20J%2Fm%5E2)
Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as
![\Delta P = \frac{4 s}{R}](https://tex.z-dn.net/?f=%5CDelta%20%20P%20%20%3D%20%20%5Cfrac%7B4%20s%7D%7BR%7D)
Now the pressure difference for the first bubble is mathematically evaluated as
![\Delta P _1 = \frac{4 s}{r_1}](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_1%20%3D%20%20%5Cfrac%7B4%20s%7D%7Br_1%7D)
substituting values
![\Delta P _1 = \frac{4 *25 *10^{-3}}{0.01}](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_1%20%3D%20%20%5Cfrac%7B4%20%2A25%20%2A10%5E%7B-3%7D%7D%7B0.01%7D)
![\Delta P _1 =10 J/m^3](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_1%20%3D10%20%20J%2Fm%5E3)
Now the pressure difference for the second bubble is mathematically evaluated as
![\Delta P _2 = \frac{4 s}{r_1}](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_2%20%3D%20%20%5Cfrac%7B4%20s%7D%7Br_1%7D)
![\Delta P _2 = \frac{4 *25 *10^{-3}}{0.005}](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_2%20%3D%20%20%5Cfrac%7B4%20%2A25%20%2A10%5E%7B-3%7D%7D%7B0.005%7D)
![\Delta P _2 =20 J/m^3](https://tex.z-dn.net/?f=%5CDelta%20%20P%20_2%20%3D20%20%20J%2Fm%5E3)
A line has negative slope if it slopes downward from left to right on a graph.
Answer:
(e)-The current is doubled.
Explanation:
According to Ohm law,
where I is the current, V is the voltage and R is the resistant.
![I_1 = \frac{V_1}{R_1}](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7BV_1%7D%7BR_1%7D)
![I_2 = \frac{V_2}{R_2}](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7BV_2%7D%7BR_2%7D)
If V is doubled,
while R remains constant
we would have
![I_2 = \frac{2V_1}{R_1} = 2(\frac{V_1}{R_1}) = 2I_1](https://tex.z-dn.net/?f=I_2%20%3D%20%5Cfrac%7B2V_1%7D%7BR_1%7D%20%3D%202%28%5Cfrac%7BV_1%7D%7BR_1%7D%29%20%3D%202I_1%20)
The current is doubled.
Due to the barrier being broken the sound will go threw but if the door was closed all the way it would be different and there would be no space for the sound to slip threw