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3 years ago

100 points!!! Please help!!!

Physics

2 answers:

sashaice [31]3 years ago

6 0

Answer:

please find the answer in the attached image, divided in 2 parts.

Phantasy [73]3 years ago

5 0

Answer:

ANSWER IS 61.6 J...

HOPE IT HELPS YOU..

I AM NOT ABLE TO ATTACH THE IMAGE ,

SORRY FOR THAT..

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A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl

Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

5 0

3 years ago

What happens at a reverse fault plane?

melamori03 [73]

<span>B. Compression causes upward vertical movement.</span>

5 0

3 years ago

Read 2 more answers

Hiii please help i’ll give brainliest if you give a correct answer please thanks!

ryzh [129]

Answer:

i Guess 1 or 3

Explanation:

yea 1 or 3 or just guess hard

7 0

3 years ago

A population has a birthrate of 10/1000 , a death rate of 9/1000, an immigration rate of 3/1000, and an emigration rate of 7/100

Charra [1.4K]

Birth rate 10/1000 = +population

Death rate 9/1000 = -population

Immigration rate 3/1000 = +population

Emigration rate 7/1000 = -population

10/1000 - 9/1000 + 3/1000 - 7/1000= -3/1000

Growth rate = (-3/1000)(100%)= -0.3%

3 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago