Answer:
The specific gravity of the unkown liquid is 15.
Explanation:
Gauge pressure, at the bottom of the tank in this case, can be calculated from
where 
and 
are the height of the column of oil and the unkown liquid, respectively. Writing for 
, we have
Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle
Answer:
Then the difference of weight between the two cars are:
Δw = 14210 - 5292 = 8918 N
Explanation:
An object's weigh due to the gravitational attraction force of the earth is:
w = mg
Where: m is the object's mass
g is the gravitational acceleration in the surface earth
g = 9.8 m/s2
The the ultralight car's weight is:
And the Honda Accord's weight is:
Then the difference of weight between the two cars are:
Δw = 14210 - 5292 = 8918 N
A motorboat accelerates uniformly from a velocity of 6.5m/s
to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2
to the east ,
how far did it travel during the accelration? Give your
answer in units of kilometers per hour/sec. To find the acceleration of the car
we have to
<span>
1. First determine
the suitable formula for this word problem.
Which is a. A=vf-vi/t</span>
which will be
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s
Solution:
<span>
x = v0t + ½at2</span>
<span>x = <span>16.645375 m </span></span>
Answer:
P₂ = 138.88 10³ Pa
Explanation:
This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation
Let's start by looking for the top speed
Q = A₁ v₁ = A₂ v₂
We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)
v₂ = A₁ / A₂ v₁
They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)
v₂ = 2 1.20
v₂ = 2.40 m / s
Now let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
Let's clear the pressure at point 2
P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)
we put our reference system at the lowest point
y₁ - y₂ = -20 cm
Let's calculate
P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)
P₂ = 143 103 - 2,160 103 - 1,960 103
P₂ = 138.88 10³ Pa
