Do 112m /29s which it will be 3.862 which if you round it, it will be 3.86 m/s
Answer:
Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters
Explanation:
First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).
Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.
The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.
Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)
Answer:
27: 85
28:75%
Explanation:
27:68=80
?=100 hence (68×100)÷80
=85
28:<em>1</em><em>8</em><em>/</em><em>2</em><em>4</em><em>×</em><em> </em><em>1</em><em>0</em><em>0</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>5</em><em>%</em>
Answer:
a=0.555m/s^2
Explanation:
First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement
X=VT
X=(18)(0.45)=8.1m
then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result
X=300-8.1=291.9
finally we use the equation for constant acceleration
Vf=0 final speed
Vo=18m/s= initial speed
X=291.9m
(Vf^2-Vo^2)/2X=a
(0-18^2)/(2*291.9)=a
a=0.555m/s^2