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elixir [45]
3 years ago
8

The volume of an object as a function of time calculated by V=At^3+B/t,where t is time measured in seconds and V is in cubic met

ers. Determine the dimension of the constants A and B
Physics
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

A = L^3 T^-3  ,  B = L^3 T

Explanation:

Given: volume ( V )  = At^3 + B/t  ------ ( 1 )

dimension of volume = L^3

and the Dimension of time = T

back to equation ( 1 )

L^3 = A * T^3   ------- ( 2 )

also L^3 = B/T ------- ( 3 )

from equation ( 2 )

A = L^3 T^-3

from equation ( 3 )

B = L^3 T

<u>The dimensions of the constants A and B </u>

A = L^3 T^-3  ,  B = L^3 T

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In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w
VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

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2 years ago
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zhannawk [14.2K]

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2 years ago
A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel
zloy xaker [14]

Answer:

4.25 m/s^{2}

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s

Change in velocity considering the y component will be

Final velocity-Initial velocity

\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s

Resultant change in velocity=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s

Acceleration= change in velocity per unit time hence

a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}

5 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
Which best describes the ages of the rocks on opposite sides of and the same distance from, A Seafloor spreading center?
Genrish500 [490]

Answer;

-The rocks are the same age

Explanation;

Seafloor spreading is the process by which the seafloor moves apart at mid-ocean ridges.  Divergent seafloor spreading occurs at this type of plate boundary.

Seafloor spreading and other tectonic activity processes are the result of mantle convection. Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle’s convection currents makes the crust more plastic and less dense.


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3 years ago
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