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kompoz [17]
3 years ago
8

A small hot-air balloon is filled with 1.01×106 l of air (d = 1.20 g/l). as the air in the balloon is heated, it expands to 1.10

×106 l. what is the density of the heated air in the balloon?
Physics
1 answer:
lara31 [8.8K]3 years ago
5 0
<span>Volume of air in the balloon 1.01 x 10^6 L
 Density of air is 1.20 g/l
 Mass = Density X Volume

 So mass of the air in the Balloon= ( 1.01 x 10^6) X 1.20 = 1.212 x 10^6 g
 As the air is heated, the volume of air in the balloon expands to 1.10x 10^6 L Density= Mass/ voume So the Density of heated air = 1.212 x 10^6/ 1.10x 10^6 = 1.101 g/l The answer is 1.101 g/l.</span>
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Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

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8 0
1 year ago
A sewing machine uses 1.6kWh of electricity in one day. If electricity costs 9p per unit, what is the total cost in pence of usi
atroni [7]

Answer:

\huge\boxed{\sf 1.6\ units = 14.4p}

Explanation:

Since,

<h3><u>1 kWh = 1 unit</u></h3>

So,

1.6 kWh = 1.6 units

If,

<h3>1 unit = 9p</h3>

1.6 units = 9p × 1.6

1.6 units = 14.4p

\rule[225]{225}{2}

8 0
1 year ago
A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
Afina-wow [57]

Answer:

the time interval that an earth observer measures is 4 seconds

Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

t = 3.2 / √( 1 - 0.36 )

t = 3.2 / √0.64

t = 3.2 / 0.8

t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

6 0
3 years ago
Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then
Nataly_w [17]

Answer:

The option (b) is correct.

Explanation:

The expression for the power in terms of work done is as follows;

P=\frac{W}{t}

Here, W is the work done, t is the time taken and P is the power.

According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.

The expression for the equivalent resistance in the series combination is as follows;

R_{eq}=R_{1}+R_{2}......+R_{6}

Put R_{1},R_{2}......,R_{6}=R.

R_{eq}=R+R+R+R+R+R

R_{eq}=6R

It is given in the problem that a seventh resistor is added (in series).

R_{eq}=R_{1}+R_{2}......+R_{7}

Put R_{1},R_{2}......,R_{7}=R.

R_{eq}=R+R+R+R+R+R+R

R_{eq}=7R

The expression for the power in terms of voltage and resistance is as follows;

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

Here, R is the resistance.

From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.

If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.

Therefore, the option (b) is correct.

5 0
3 years ago
Light from a single laser is directed through two slits that are separated by a small distance. On the other side of the slits i
nekit [7.7K]
The answer is destructive interference.  You have this for both C and D.  I suspect one of C or D is supposed to be constructive interference...  But destructive interference is the answer
3 0
3 years ago
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