Answer:
3311N
Explanation:
r = radius = 600m
V = speed = 150m/s
Mass = weight = 70kg
The weight of pilot when calculated due to circular motion
W = tv
Fv = mv²/r
Fv = 70x150²/600
Fv = 79x22500/600
= 15750000/600
= 2625N
Real Weight of the pilot = m x g
= 70 x 9.8
= 686N
The apparent Weight is calculated by
Mv²/r + mg
= 2625N + 686N
= 3311 N
Therefore the apparent Weight is 3311N
Answer:
The pressure at this point is 0.875 mPa
Explanation:
Given that,
Flow energy = 124 L/min
Boundary to system P= 108.5 kJ/min
We need to calculate the pressure at this point
Using formula of pressure
Here, 
Where, v = velocity
Put the value into the formula
Hence, The pressure at this point is 0.875 mPa
answer;
The hole in the center of the washer will expand
explanation;
<em>A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.</em>
this is an example of area expansivity.
coefficient of area expansivity is change in area per area per degree rise in temperature
a=dA/(A*dt)
as the temperature rises , there will be volumetric and area expansivity on the body. volume also increases because of the intermolecular forces of attraction between the molecule is now getting apart.
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
