∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)
2nd and only 2nd option is right
Answer:
Magnification will be equal to 3
Explanation:
We have given focal length of the converging lens 
Focal length of the diverging lens 
Object is placed 40 cm to the length of the converging lens d = 40 cm
Combination of the focal length will be equal to


F = 60 cm
So combination of the focal length will be 60 cm
Magnification is given by

So magnification will be equal to 3
The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer:
The time rate of change in air density during expiration is 0.01003kg/m³-s
Explanation:
Given that,
Lung total capacity V = 6000mL = 6 × 10⁻³m³
Air density p = 1.225kg/m³
diameter of the trachea is 18mm = 0.018m
Velocity v = 20cm/s = 0.20m/s
dv /dt = -100mL/s (volume rate decrease)
= 10⁻⁴m³/s
Area for trachea =

0 - p × Area for trachea =



⇒

ds/dt = 0.01003kg/m³-s
Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s