Please help quick please its a test

a balloon inflated in a room at 297k has a volume of 4.00 l. the balloon is then heated to a temperature of 331 k. what is the n

Naddika [18.5K]

V2 = 4.4579 L

Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2

4.00 L = V2
———- ———
297 K 331 K

Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)

V2 = 4.4579 L

Round to significant digits if required

7 0

2 years ago

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

7 0

3 years ago

What is the frequency of a wave?

Paladinen [302]

The answer should be the number, or amount, of cycles that occur in a given time. Frequency is basically Hertz. Frequency is the number of waves that pass through a certain point every second. I hope that I was able to answer your question.
Peace out!

- Hershy103

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3 years ago

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A sound wave traveling eastward through air<span> causes the air molecules to Vibrate east and west. </span>

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Explain Why a flying aeroplane has more Kinetic Energy than a flying insect?

Airida [17]

Answer:

Why do insects fly so high?

Because the angle of attack is so high, a lot of momentum is transferred downward into the flow. These two features create a large amount of lift force as well as some additional drag. The important feature, however, is the lift.

Why an Aeroplane flying has kinetic

A flying aeroplane has potential energy has it flies above the ground level. And since the aeroplane is flying motion is associated with it and thus possesses kinetic energy. Hence a flying aeroplane has both potential and kinetic energ

Explanation:

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3 years ago