Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
FORMULA:
- V = IR, where V = P.D; I = Current; R = Resistance.
ANSWER:
Total equivalent resistance for circuit:
R(eq) = R1 + R2 [It is in series]
Now, Current passing through whole circuit:
I = V/R
We know that, In series combination current passing through whole circuit is same.
So, V¹ = IR¹
V¹ = 1/50 × 330
And V² = IR²
V² = 1/50 × 470
The final speed of the car at the given conditions is 30.1 m/s.
The given parameters:
- <em>Mass of the car, m = 1700 kg</em>
- <em>Velocity of the car, v = 21 m/s</em>
- <em>Time of motion, t = 10 s</em>
- <em>Additional energy provided by the engine, E₁ = 22,000 J</em>
- <em>Energy used in overcoming friction, E₂ = 3,666.67 J</em>
The change in the energy applied to the car is calculated as;
The final speed of the car is calculated as follows;
Thus, the final speed of the car at the given conditions is 30.1 m/s.
Learn more about change in kinetic energy here: brainly.com/question/6480366
Answer:
See below
Explanation:
At point A the PE = mgh = 2 * 10 * 1 = 20 J
at point B, all of the PE , 20 J , is converted to Kinetic Energy
KE = 1/2 m v^2
20 = 1/2 (2)(v^2 )
20 = v^2 v = sqrt 20 = 4.47 m/s
for the friction part
vf = vo t + 1/2 a t^2 vf = final velocity = 0 (stopped)
vo = original velocity = 4.47 m/s
a = -1 m/s^2
0 = 4.47 t + 1/2 (-1) t^2
- .5t^2 + 4.47 t = 0
t ( -.5t+ 4.47) = 0 shows t = 4.47/.5 = 8.9 seconds
