Answer:
<h2>4900 J</h2>
Explanation:
The gravitational potential energy of a body can be found by using the formula
GPE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
GPE = 5 × 9.8 × 100
We have the final answer as
<h3>4900 J</h3>
Hope this helps you
1) Try to head into the waves at some slight angle and the speed of the boat should be reduced.
2) In order to ride up and over the waves, the speed of the boat should be slow.
3) The less the speed of the boat, and the less strain will be put on the hull and superstructure.
Answer:
20m/s^2
Explanation:
Acceleration=Change in velocity/time taken for change
40-20/1
20m/s^2
Answer: The elevator must be accelerating.
Explanation:
As the tension force is opposing to the the force of gravity on the load which is hung vertically, and the tension force can adopt any value in order to comply with Newton's 2nd law, if the tension force is less than the force due to gravity, this means that all system is not in equilibrium, so it must be accelerating.
If we assume that the downward is the positive direction, we can write:
mg - T = ma
If T = 0.9 mg, ⇒ mg (1-0.9) =0.1 mg = m a ⇒a = 0.1 g , in downward direction.
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