Answer:
<h3>Hey there ! </h3><h3>Here is your Answer Buddy | </h3>
<h3>• Define Instance :- </h3>
- The data stored in database at a particular moment of time is called instance of database.
<h3>• Define Schema :- </h3>
- Database schema defines the variable declarations in tables that belong to a particular database
<h3><u>The value of these variables at a moment of time is called the instance of that database.</u></h3><h3 />
Explanation:
<h3>Hope this helps !! </h3>
<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
Explanation:
<h2>Given:</h2>
m = 1,250 kg
= 30 km/hr
F = 1,700 N
t = 10 s
<h2>Required:</h2>
Final velocity
<h2>Equation:</h2><h3>Force</h3>
F = ma
where: F - force
m - mass
a - acceleration
<h3>Acceleration</h3>
a = 
where: a - acceleration
- initial velocity
- final velocity
t - time elapsed
<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>
F = ma
Substitute the value of F and m
(1700 N) = (1250 kg)(a)
a = 
a = 1.36 m/s²
<h3>Solve for final velocity using the formula for acceleration</h3>
= 
= 
- Substitute the value of a,
and t
a = 


= 
= 
<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
Answer:
Hello your question is incomplete attached below is the complete question
answer :
U/r = R =
[ 1 -
] = 0
∴ No slip condition is satisfied
Explanation:
Given that
Outlet velocity U(r) =
[ 1 -
]
<u>prove that the output velocity profile satisfies the no slip condition</u>
at no slip u = 0 ( i.e. at the pipe's inner surface )
at , r = R ( inference is at center )
hence U/r = R =
[ 1 -
] = 0
∴ No slip condition is satisfied
Answer:
The force exerted by the water on one side of the plate is F = 24*pg
Explanation:
From the given question, the first step to take is to find he force exerted by the water on one side of the plate.
Solution
Given that:
Let the pressure the at a depth of y ft be = pgy lb/Pa
the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)
Then
we combine with the range for y as = y E [2 , 5]
Thus,
F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy
Recall that,
p = water density
g= gravity of acceleration
so,
F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]
F = 4/3*pg * [y^3/3 - y^2] [2 , 5]
F = 4/3*pg * [18]
Finally, F = 24*pg