Answer:
The tension in the tow rope pulling the log is 784.61 N
Explanation:
Given:
v = speed = 13 m/s
Pf = final power when the log is pulled = 8.5x10⁴W
Pi = average power = 7.48x10⁴W
The force required to move a car is equal to:
T = Ff - Fi = 6538.46 - 5753.85 = 784.61 N
Answer:
3136
Explanation:
because you multiply 780 by 4 because if it has a constant speed meaning it will not change speed then you multiply it by 4 because it says in 4 hours ( I hope this helped)
Answer:
83.3 Wb
Explanation:
The magnetic flux linkage through the coil is given by:
where
B is the magnetic field strength
A is the cross sectional area
N is the number of turns in the coil
is the angle between the direction of the field and the normal to the coil
In this problem:
B = 1.74 T
A = 0.133 m^2
N = 360
Therefore, the magnetic flux linkage is
Answer:
Proper Inflation and Feel When inflating your ball, you can use either a hand pump or an air pump equipped with a gauge that gives readings in pounds per square inch, also called psi. Footballs used in the NFL are inflated to 13 psi, but a proper range can fall between 12.5 and 13.5 psi, according to Wilson Sporting Goods.
Explanation:
Answer:
2.5 * 10^-3
Explanation:
<u>solution:</u>
The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:
<em>δu/δx+δv/δy=0</em>
so that:
<em>δv/δy= -δu/δx</em>
Now, since u = Uy/δ, where δ = cx^1/2, we have that:
<em>u=U*y/cx^1/2</em>
and we obtain:
<em>δv/δy=U*y/2cx^3/2</em>
The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):
v=∫δv/δy(dy)=U*y/4cx^1/2
=y/x*(U*y/4cx^1/2)
=u*y/4x
which is exactly what we needed to demonstrate.
Also, using u = U*y/δ in the last equation we can obtain:
v/U=u*y/4*U*x
=y^2/4*δ*x
which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:
(v/U)_max=δ^2/4δx
=δ/4x
=2.5 * 10^-3
