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4 years ago

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.

Physics

1 answer:

Andreas93 [3]4 years ago

4 0

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

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Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

$P=VI$

Express this in terms of 'I'. This gives,

$I=\frac{P}{V}$

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, $P_1=16\ W, V=110\ V$

So, current drawn is given as:

$I_1=\frac{16\ W}{110\ V}=0.145\ A$

For incandescent bulb, $P_2=100\ W, V=110\ V$

So, current drawn is given as:

$I_2=\frac{100\ W}{110\ V}=0.909\ A$

For CFL bulb, $P_3=24\ W, V=110\ V$

So, current drawn is given as:

$I_3=\frac{24\ W}{110\ V}=0.218\ A$

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

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3 years ago

A comparison of a machine’s work output with its work input is called __??__.

Aloiza [94]

Ohhhhh its called a input machine

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3 years ago

A cylindrical piece of aluminum is 6.00cm y’all and 2.00cm in radius how much does it weigh

Harlamova29_29 [7]

Answer:

2.00 N

Explanation:

Weight is mass times gravity:

W = mg

Mass is density times volume:

m = ρV

Volume of a cylinder is:

V = πr²h

Finding the volume:

V = π (2.00 cm)² (6.00 cm)

V = 75.4 cm³

The density of aluminum is 2.7 g/cm³. Finding the mass:

m = (2.7 g/cm³) (75.4 cm³)

m = 204 g

Finding the weight:

W = (0.204 kg) (9.8 m/s²)

W = 2.00 N

The aluminum cylinder weighs 2.00 N.

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3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

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3 years ago

A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the ground,

katrin [286]

Answer:

0.405 seconds

Explanation:

Consider the amount of time it takes the block to fall from 53 m up to 14 m above the ground; then consider the amount of time it takes the block to fall from 53 m up to 2 m above the ground.

First, d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 × 39 / 9.8)^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters...

d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 * 51/ 9.8 )^1/2 = 3.2262 seconds

So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

3.2262 - 2.8212 = 0.405 seconds

this is how much time there is from when the man sees the block until it hits him. Not much time...

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3 years ago

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.