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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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The answer is "83.1%".

Explanation:

Given:

$\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}$

Using formula:

$\to g_E = G \frac{M_E}{(R_E +h)^2}$

$\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\$

$\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}$

Calculating the gravity on the Earth’s surface:

$\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}$ $= \frac{8.15}{9.8} \times 100=83.1 \%$

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Given that acceleration of the airplane is

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$v_y = -15.5 m/s$

now we have displacement in x direction given as

$x = v_x t + \frac{1}{2}a_x t^2$

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$d = \sqrt{2002^2 + 2961^2}$

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Define the term dimension​

Define the term dimension