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3 years ago

Compare the densities of two objects that have the same volume, but one feels heavier than the other.

1 answer:

8 0

We know, density = Mass / Volume
It represents density is directly proportional to mass and indirectly proportional to volume. So, at a constant volume, object with larger density will appear more heavier than that of object with smaller density.

In short, Heavier object will have the higher density than the other.

Hope this helps!

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Katrina inserts a key into her car’s ignition. She turns the key, putting her car into ignition mode. In this mode, some energy

Anestetic [448]

Answer: The option (C) is correct.

Explanation:

Ignition is the process in which the fuel is made to start burn. In car's ignition, it is the part where the key is turned so that the engine may get start.

The fossil fuel which is used in the car's engine is petroleum. Therefore, the chemical energy is used to allow the fuel to burn and move the car.

4 0

3 years ago

Read 2 more answers

You connect a 250-2 resistor, a 1.20-mH inductor, and a 1.80-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. When

Basile [38]

Answer:

Explanation:

reactance of inductor = wL = 2 X 3.14 X 60 X 1.2 X 10⁻³ = .45 ohm.

reactance of capacitor = 1/wC = 1/( 2 X 3.14 X 60 X 1.8 X 10⁻⁶ ) = 1474.4

Impedence of the circuit =[ R² + ( I/wC - wL) ]¹/² = [250² + ( 1474.4-.45 )]¹/²

Impedence = 1495 ohm.

RMS Voltage = 120/ 1.414 = 84.86 V

current = 84.86 / 1495 = 0.0576

Potential over resistance = 0.0576 x 250 = 14.2 V.

4 0

3 years ago

I would like to know the answers to both 10 & 11

oksian1 [2.3K]

Answer:

picture is not showing up sorry

Explanation:

7 0

4 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time <em>t</em> :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time <em>t</em> :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$