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nekit [7.7K]
3 years ago
11

Avg velocity definition | Problem Average velocity vectors…

Physics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

Explanation:

As we know that average velocity is defined as the ratio of change in position and the time interval of the object

It is given as

v_{avg} = \frac{\Delta x}{\Delta t}

so we will have

\Delta x = displacement

\Delta t = time interval

the direction is this average velocity is always along the direction of the displacement

so correct answer will be

point in the same direction as the change in position.

require knowing the initial and final position, along with the elapsed time.

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SCORPION-xisa [38]

Answer:

a

Explanation:

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3 years ago
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8 0
3 years ago
Determine the amount of potential energy of a 5N book that is 1.5m high on a shelf.
Alex73 [517]

Answer:

Potential energy of book = 7.5 J

Explanation:

Given:

Weight of book = 5 N

Height of shelf = 1.5 meter

Find:

Potential energy of book

Computation:

Weight = Mass x Acceleration of gravity

Mass x Acceleration of gravity = 5 N

Potential energy = Mass x Acceleration of gravity x Height

Potential energy of book = Mass x Acceleration of gravity x Height

We know that;

Mass x Acceleration of gravity = 5 N

So,

Potential energy of book = 5 x 1.5

Potential energy of book = 7.5 J

3 0
3 years ago
Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e
My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

PV =NkT

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

N = 1

k = 1.38*10^{-23}J/K

T = 27\°C = 27\°C + 273 = 300K

P = 1atm = 101325Pa

Rearrange the equation to find V we have,

V = \frac{NkT}{P}

V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}

V = 4.0858*10^{-26}m^3

We know that length of a cube is given by

V = L^3

Therefore the Length would be given as,

L = V^{1/3}

L = (4.0858*10^{-26})^{1/3}

L = 3.445*10^{-9}m

Therefore each length of the cube is 3.44nm

7 0
3 years ago
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

B =  \frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}  

solve we get

B = 0.692 T

8 0
3 years ago
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