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malfutka [58]
3 years ago
15

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Velocity:

Velocity is change in displacement with respect to time:

\frac{\Delta x}{\Delta t}

Analysing the units, meters (displacement) and seconds (time) are basic units:

\frac{m}{s}

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

F = ma

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

a=\frac{\Delta v}{\Delta t}

Analysing the units:

\frac{\frac{m}{s}}{s}=\frac{m}{s^2}

Therefore, the unit of force is:

kg\frac{m}{s^2}

Pressure:

Pressure is given by the equation:

P=\frac{F}{S} where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

[l^2]=m^2, the unit of area

Dividing the aforementioned unit of force by the unit of area:

\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}, the unit of pressure

Work:

Work is given by the equation:

W=\vec{F}\cdot \vec{x}, (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}, the unit of work.

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A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
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- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
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ANTONII [103]

Answer:

6.0 N

Explanation:

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The formula to apply here is :

Force= mass * acceleration

F=ma

Mass, m = 4 kg

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The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb
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Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

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Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

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Explanation:

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