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4 years ago

A reconnaissance plane flies 605 km away from

Physics

1 answer:

kolezko [41]4 years ago

8 0

Answer:

$v_{avg} =$ 355 m/s

Explanation:

Distance = 605 km

Initial speed = $v_{i}$ = 284 m/s

Final velocity = $v_{f}$ = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

$2as = v_{f}^{2}+v_{i}^{2}$

Then using first equation of motion, we find time

$v_{f} = v_{i}+at$

Then using the formula of average velocity, we find average velocity

$v_{avg}=\frac{total-distance}{total-time}$

Second method is very simple

$v_{avg}=\frac{v_{f}+v_{i} }{2}$

$v_{avg}=\frac{426+284}{2}$

$v_{avg} =$ 355 m/s

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3 years ago

Read 2 more answers

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

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3 years ago

Can someone please help me with science.

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Answer:

The answers to your questions are given below

Explanation:

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f => is the frequency

From the equation i.e E = hf, we can conclude that the energy of a wave is directly proportional to it's frequency. This implies that an increase in the frequency of the wave will lead to an increase in the energy of the wave and also, a decrease in the frequency will lead to a decrease in the energy of the wave.

23. Gamma ray and radio wave are both electromagnetic waves. All electromagnetic waves has a constant speed of 3×10⁸ m/s in space.

Thus, gamma ray and radio wave have the same speed in space.

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3 years ago

In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-

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Answer:

a) truc is C, b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

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v = 3 109 m / 20

v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time t = to γ

For distance L = Lo / γ

γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time t = 20 1.15 = 23 s

The distance L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0

4 years ago

Two 10cm diameter metal disks separated by a 0.63mm thick piece of pyrex glass are charged to a potential difference of 1000V. D

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Parallel-plate capacitor has there fore formula is

C=(ϵ0A)/d
putting valuesC=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F then Q=CV=1.1*10^-10*1000=1.1*10^-7C
as
η=Q/Atherefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
hope this helps

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3 years ago