At 100 km/hr, the car's kinetic energy is
KE = (1/2) (mass) (speed)²
KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²
KE = (787.5 kg) (27.78 m/s)²
KE = 607,639 Joules
In order to deliver this energy in 2.9 seconds, the engine must supply
(607,639 J / 2.9 sec) = 209,531 watts
<em>Power = 281 HP</em>
Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Explanation:
The changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is
We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.
