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2 years ago

Define the term dimension

Physics

1 answer:

Vladimir79 [104]2 years ago

4 0

Answer:

Q1. A measurable extent of a particular kind, such as length, breadth, depth, or height.

Q2. A dimensional constant is a physical quantity that has dimensions and has a fixed value. Some of the examples of the dimensional constant are Planck's constant, gravitational constant, and so on.

Q3. Physical quantities which posses dimensions and have variable are called dimensional variables. Examples are length, velocity, and acceleration etc.

Q4. Dimensionless variables are the quantities which doesn't have any dimensions the the value is a variable. Eg: angle = arc/ radius. Dimensions = L/L. = 1. So angle does not have any dimensions and the value can vary.

Q5. Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Q6. Dimensional analysis has been around a long time, Newton called it the "Great principle of Similitude", but the modern form can be traced back to James Clerk Maxwell. It was Maxwell who distinguished mass [A/], length [£], and time [7"] as the independent dimensions from which others could be derived.

Q7. Mass, length, time, temperature, electric current, amount of light, and amount of matter.

Q8. Dimensional analysis is used to convert the value of a physical quantity from one system of units to another system of units. Dimensional analysis is used to represent the nature of physical quantity. The expressions of dimensions can be manipulated as algebraic quantities.

Hope that helps. x

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Explanation:

From the question we are told that

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The surface area is $A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2$

The height of fall is h =4.8 m

The length of the prism is $l = 23.2 = 0.232 \ m$

The width of the prism is $w = 11.6 = 0.116 \ m$

The terminal velocity is mathematically represented as

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$C$ is the drag coefficient for a horizontal skydiver with a value = 1

A is the area of the prism the squirrel is assumed to be which is mathematically represented as

$A = 0.116 * 0.232$

$A = 0.026912 \ m^2$

substituting values

$v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }$

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3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

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(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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2 years ago

Define the term dimension​

Define the term dimension