Best Answer:<span> </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km
The work done by the aircraft overcoming the wind is
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
Answer:
160.75 N
Explanation:
The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.
As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.
When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:
F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N
The coefficient of kinetic friction<span> is the force between two objects when one object is moving, or if two objects are moving against each other</span>
The power is 833.3 W
Explanation:
First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

where
mg = 1250 N is the weight of the barbell
h = 2 m is the change in height
Substituting,

Now we can calculate the power, which is equal to the work done per unit time:

where
W = 2500 J is the work done
t = 3 s is the time taken
Substituting,

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