Answer:
The gravitational force on the elevator = 4500N
Explanation:
The given parameters are;
The force applied by the elevator, F = 4500 N
The acceleration of the elevator = Not accelerating
From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows;
F = W + Mass of elevator × Acceleration of elevator
∴ F = W + Mass of elevator × 0 = W
F = 4500 N = W
The net force on the elevator is F - W = 0
The gravitational force on the elevator = W = 4500N.
The gravitational force on the elevator, if the elevator is NOT accelerating is 4500 N.
Given data:
The magnitude of force on the elevator is, F = 4500 N.
From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows,
F = W + ma
Here, a is the acceleration of the elevator. Since, the elevator is not accelerating. Then, a = 0 .
The weight of the elevator is due to the gravitational force. So, we need to obtain the weight as gravitational force.
Solving as,
[tex]F = W+ma\\\\F=W +(m \times 0)\\\\4500=W\\\\W = 4500 \;\rm N[/tex]
Thus, we can conclude that the gravitational force on the elevator, if the elevator is NOT accelerating is 4500 N.
Learn more Newton's third law here:
https://brainly.com/question/17529892
An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine:
(a) the volume of the tank
(b) the final temperature
(c) the electric power rating of the resistor
Solution:
Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg
Temperature = 200 °C
And 25% of the volume by liquid water is steam.
State 1
[tex]$m=\frac{V}{v}$[/tex]
[tex]$m=m_f+m_g$[/tex]
[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]
[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex] (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C )
[tex]$V=6.304 \times 10^{-3}$[/tex]
Now quality of vapor
[tex]$x=\frac{m_g}{m}$[/tex]
[tex]$=3.377 \times 10^{-3}$[/tex]
Internal energy at state 1 can be found out by
[tex]$u_1=u_f+xu_{fg}$[/tex]
[tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]
= 856.54 kJ/kg
After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]
Tank is rigid, so volume of tank is constant.
[tex]$v_g=v_2=\frac{V}{m}$[/tex]
[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]
[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]
Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature
[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]
= 369.11° C
Internal energy at state 2
[tex]$u_2=2154.9 \ kJ/kg$[/tex]
Now power rating of the resistor
[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]
[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]
= 1.51 kW
Which of the following is a characteristic of a base?
A. pH greater than 7
B. pH between 4 and 13
C. pH between 4 and 9
D. pH less than 7
Answer:
A. pH greater than 7
Explanation:
Because a base has a pH scale of 8 to 14