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3 years ago

How are typhoons formed? Need help!

Physics

2 answers:

yan [13]3 years ago

8 0

Answer:

typhoons are formed by the people cant judge or observe

Explanation:

mark me brainliest

Mila [183]3 years ago

5 0

Answer:

typhoon forms when winds blow into areas of the ocean where the water is warm. These winds collect moisture and rise, while colder air moves in below. This creates pressure, which causes the winds to move very quickly. The winds rotate, or spin, around a center called an eye.

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Air expands isentropically from 2.2 MPa and 77°C to 0.4 MPa. Calculate the ratio of the initial to the final speed of sound.

djyliett [7]

Answer:

The ratio of initial to final speed of sound is given as 1.28.

Explanation:

As per the thermodynamic relation of isentropic expansion

$\frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}}$

Here

$P_1$ is the pressure at point 1 which is given as 2.2 MPa
$T_1$ is the temperature at point 1 which is given as 77 °C or 273+77=350K

$P_2$ is the pressure at point 1 which is given as 0.4 MPa
$T_2$ is the temperature at point 2 which is to be calculated
k is the ratio of specific heats given as 1.4

Substituting values in the equation

$\frac{T_2}{350}=(\frac{0.4}{2.2})^{\frac{1.4-1}{1.4}}\\\frac{T_2}{350}=(0.18)^{0.2857}\\T_2=(0.18)^{0.2857} \times 350 \\T_2=0.61266 \times 350\\T_2=214.43 K$

As speed of sound c is given as

$c=\sqrt{kRT}$

for initial to final values it is given as

$\frac{c_i}{c_f}=\frac{\sqrt{k_1R_1T_1}}{\sqrt{k_2R_2T_2}}$

As values of k and R is constant so the ratio is given as

$\frac{c_i}{c_f}=\sqrt{\frac{T_1}{T_2}}$

Substituting values give

$\frac{c_i}{c_f}=\sqrt{\frac{350}{214.43}}\\\frac{c_i}{c_f}=\sqrt{1.63}}\\\frac{c_i}{c_f}=1.277 \approx 1.28$

So the ratio of initial to final speed of sound is 1.28.

5 0

3 years ago

HELP ME PLEASEEEEEEEEEEEEEE

Lelu [443]

Answer: choose a

Explanation:

7 0

3 years ago

Read 2 more answers

A light spring of constant 163 N/m rests vertically on the bottom of a large beaker of water.

LUCKY_DIMON [66]

Answer:

24.71cm

Explanation:

We approach this problem base don Hooke's law which states the elongation produced in an elastic material is proportional to the applied load or force provided that its elastic limit is not exceeded. This is expressed mathematically as follows;

$F=ke................(1)$

where F is the applied force, k is the force constant and e is the elongation or extension of the material.

In this problem, the applied force F is the weight of the wood which is calculated as follows,

$F=mg.............(2)$

m = 4.11kg

$g=9.8m/s^2$

Hence,

$F=4.11*9.8\\F=40.278N$

Given that k = 163N/m, we make appropriate substitutions into equation (1) to obtain the following;

$40.278=163*e\\e=\frac{40.78}{163}\\e=0.2471m$

Since it is required in cm, we perform the conversion as follows, knowing that 100cm = 1m

$0.2471m=0.2471*100cm=24.71cm$

NB: We do not necessarily need the the density of the wood to perform our calculations since other parameters were given from which we were able to obtain its weight.

7 0

4 years ago

you measure or observe two events to occur at the same time one nearby and one further away. Which one did you actually see firs

mihalych1998 [28]

Your eyes and brain would first register the event nearby sooner than the one farther away

8 0

3 years ago

Read 2 more answers

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

1. Part A: 648N
2. Part B: 324J
3. Part C: 1,296N

Explanation:

1. Part A:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

3. Part C

What maximum force must you apply to move the platform to the position in Part B?

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

How are typhoons formed? Need help! ​

How are typhoons formed? Need help!