I REALLY NEED HELP ON THIS PLEASE HELP ME

How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2

soldier1979 [14.2K]

∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

3 0

3 years ago

Read 2 more answers

The work that a force does by acting on an object is equal to what?

maw [93]

2nd and only 2nd option is right

3 0

3 years ago

A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I

scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens $F_1=20cm$

Focal length of the diverging lens $F_2=30cm$

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

$\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}$

$\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

$M=\frac{F}{F-d}=\frac{60}{60-40}=3$

So magnification will be equal to 3

3 0

3 years ago

A car with a mass of 1.50x10^3 kg starts from rest and accelerates to a speed of 18.0m/s in 12.0 s. assume that the force of res

Luden [163]

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

8 0

3 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

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