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2 years ago

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Town A lies 15 km north of town B. Town C lies 10 km west of town A. A small plane flies directly from town B to town C. What is

the displacement of the plane?

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer:

the correct answer is b

Explanation:

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10) T F Two unequal weights are connected by a massless string which passes over a frictionless pulley. If the pulley has no app

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Answer:

2

Explanation:

2

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you start your bicycle ride at the top of a hill. you coast down the hill at a constant acceleration of 2.00 m/s^2. When you get

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198 meters I think, I could be wrong

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Energy can be changed from one form to another. Changes in the form of energy are called energy conversions. In an automobile en

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Answer: D <u>(chemical</u> -> <u>heat</u> -> <u>mechanical</u>)

In automobile engines the petrol/diesel fuel enter in to the engine cylinder, due to spark at the end of the compression, fuel burnt increase the temperature and pressure, develops heat <em>(chemical energy -> heat energy). </em><em>This heat energy acts on a piston develops the work on the crankshaft </em><em>( Heat energy -> Mechanical energy)</em><em>. </em>

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2 years ago

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– A man walks 50 meters due west, then 50 meters 60 north of west. Determine the magnitude and direction of the resultant disp

Fynjy0 [20]

Answer:

86.6, 45°

Explanation:

The diagram explains better.

Using vector component method:

We find the x and y components of the vectors :

For the first:

A = -50cos(0)i + 50sin(0)j

A = -50i

For the second:

B = -50cos(60)i + 50sin(60)j

B = -25i + 43.3j

The resultant vector is :

R = A + B

R = -50i - 25i + 43.3j

R = -75i + 43.3j

The magnitude is:

R = [(-75)² + (43.3)²]^½

R = 86.6m

The angle is

tanθ = (50/50) = 1

θ = 45°

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Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

or

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

2 years ago