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2 years ago

Has anyone heard from joshuasalazar697?? PLEASE LET ME KNOW ASAP I NEED TO TALK TO HIM ITS REALLY IMPORTANT!!

Physics

1 answer:

Ipatiy [6.2K]2 years ago

8 0

Answer:

no ma'am ill help you look

Explanation:

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is

erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P = P₀ + ρgh

P = 101325 Pa + (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0

3 years ago

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between

Sunny_sXe [5.5K]

Answer : $4.483\times 10^{15}\ m/s^2$ .

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$ ...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

$a=0.4483\times 10^{16}\ m/s^2$

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$ .

Hence, it is the required solution.

7 0

3 years ago

A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

What is the highest and lowest frequency a human being can hear

nata0808 [166]

Every person is different. But for a planet-wide overall average that roughly represents all human beings on Earth, the figures usually used are:

from 20 Hz to 20,000 Hz .

6 0

3 years ago