Name and explain two sleeping disorders

A fixed mass of an ideal gas has a volume of 800 cm3 under certain conditions. The pressure (in kPa) and temperature (in K) are

Mashutka [201]

Answer:

800cm3

Explanation:

This is a general gas law question

Which has the relationship P1V1/T1 = P2V2/T2

P1= P1 in kPa

T1 = T1 in K

V1= 800cm3

V2=?

P2= 2* P1

T2= 2*T1

The volume of gas after the changes (V2), making it as the subject of formula

V2= P1*V1*T2/P2*T1

V2 = P1 *800* 2T1 / 2P1 * T1 ; dividing accordingly, we have

V2 = 800cm3

3 0

3 years ago

. A force of 3.0 N acts through a distance of 12 m in

musickatia [10]

Answer:

<h3>The answer is 36 J</h3>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 3 × 12

We have the final answer as

Hope this helps you

6 0

3 years ago

A 20 ohm lamp and a 5 ohm lamp are connected in series and placed across a potential difference of 50 V.

iogann1982 [59]

Hi there!

1.

Since the two resistors are in series, we can simply add:
$R_T = R_1 + R_2 + ... R_n$

$R_T = 20 + 5 = \boxed{25 \Omega}$

2.

In series, the potential difference of each resistor (lamp) ADDS UP. We can begin by finding the current through the circuit using Ohm's law:
$V = IR\\\\I = \frac{V}{R_T}$

Plug in the values:
$I = \frac{50}{25} = 2 A$

Now,

we can use Ohm's law to find the individual voltage for each lamp.

20 Ohm lamp:
$V = 2 * 20 = \boxed{40 V}$

5 Ohm lamp:
$V = 2 * 5 = \boxed{10 V}$

3.

To solve, we can use the power equation.

$P (\text{Watts})= IV$

Plug in the values for each.

20 Ohm lamp:
$P = 2 * 40 = \boxed{80 W}$

5 Ohm lamp:
$P = 2 * 10 = \boxed{100 W}$

7 0

2 years ago

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

3 years ago

What are neglible weights. Definition?

Alex

Answer:

Negligible weights is a change so minor or insignificant to be deemed to have no effect on weight or balance.

4 0

2 years ago