Answer:
Explanation:
Mass =11.2kg
Constant velocity =3.3m/s
μk=0.25
Since the body is moving in constant velocity, then the acceleration is zero(0).
ΣF = Σ(ma)
The normal force acting on the body is upward and the weight is acting downward
Then ΣFy=0
Therefore, N=W
W=mg=11.2×9.8=109.76N
So, N=W=109.76N
Frictional force is given as
Fr=μkN
Fr=0.25×109.76
Fr=27.44N
Frictional force acting against the motion is 27.44N
Then the forward force moving the body forward
ΣF = Σ(ma)
Since a = 0
Then,
ΣF = 0
F-Fr=0
Then F=Fr
So the force moving the body forward is 27.44N
Answer:
12.6 (3 sig. fig.)
Explanation:
(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)
The initial temperature of the bar is 25. To get to the t temperature you need to add (t-25) degrees Celsius.
for 1 degree................... 7 Joules
y given degree........ p Joules
p=7y
In our case y=(t-25) .
h(t) = 7(t-25) which is the final answer.
I don’t know what book you’re talking about so I can’t help but have a look online, you may be able to find it if you search up the book name and look around a few websites
<h2>
Answer:442758.96N</h2>
Explanation:
This problem is solved using Bernoulli's equation.
Let 
be the pressure at a point.
Let 
be the density fluid at a point.
Let 
be the velocity of fluid at a point.
Bernoulli's equation states that 
for all points.
Lets apply the equation of a point just above the wing and to point just below the wing.
Let 
be the pressure of a point just above the wing.
Let 
be the pressure of a point just below the wing.
Since the aeroplane wing is flat,the heights of both the points are same.
So,
Force is given by the product of pressure difference and area.
Given that area is 
.
So,lifting force is 
