Ask question

Ask question

All categories

3 years ago

9

An airplane of mass 39,043.01 flies horizontally at an altitude of 9.2 km with a constant speed of 335 m/s relative to Earth. Wh

at is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the plane? Give your answer in scientific notation.

1 answer:

Likurg_2 [28]3 years ago

8 0

Answer:

1.2 x 10¹¹ kgm²/s

Explanation:

m = mass of the airplane = 39043.01

r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m

v = speed of airplane = 335 m/s

L = Angular momentum of airplane

Angular momentum of airplane is given as

L = m v r

Inserting the values

L = (39043.01 ) (335) (9200)

L = (39043.01 ) (3082000)

L = 1.2 x 10¹¹ kgm²/s

You might be interested in

Một vôn kế đang đo hiệu điện thế ở thang đo có giá trị mỗi độ chia là 0,1 V. Nếu kim chỉ thị nằm giữa vạch 5,5 V và 5,6 V thì đọ

kakasveta [241]

Answer:

5.6V nha bạn

Explanation:

5 0

3 years ago

Plz someone help me Asap

-Dominant- [34]

Answer:

all I know is C

are there more questions? anyone?

-KARL IS STOOPID

Explanation:

6 0

3 years ago

A piston is filled with gas. When the pressure is increasing, what is happening to the volume? Assume that all other properties

Kryger [21]

The correct answer is letter C. Volume is decreasing. For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume<span> are </span>inversely proportional<span>. </span>

3 0

3 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

3 years ago

The Earth's gravity keeps us from flying out to space. How can you explain that we have a stronger gravitational attraction to t

julia-pushkina [17]

It's the fourth choice.

This is because, since we are closer to the Earth, the Earth will have a stronger gravitational pull on us since again, we are closer.
That also explains tides, but that's just getting off topic. Hope I helped.

6 0

3 years ago

Read 2 more answers