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wariber [46]
3 years ago
10

Different organisms live in different types of ecosystems because they are adapted to different conditions. true or false

Physics
1 answer:
Marizza181 [45]3 years ago
6 0
True! Some organisms are built for different ecosystems, climates, etc. and some choose to adapt to the living conditions.

I hope that helped!!
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Which should be done in case of a laboratory accident?
stich3 [128]
Tell your instructor or teacher
4 0
3 years ago
We observe a glowing cloud of gas in space with a spectroscope. We note that many of the familiar lines of hydrogen that we know
konstantin123 [22]

Answer:

No information can be concluded from this observation alone the cloud is much cooler than hydrogen on Earth

4 0
3 years ago
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant
NNADVOKAT [17]

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

\frac{d\theta}{dt} = w_0e^{-\sigma t}

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}

Now, we can solve for sigma using the other given condition:

2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}

c.

The angular displacement is the INTEGRAL of the angular velocity function.

\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\

\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.

\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}

\theta = 8.471 rad

Convert this to rev:

8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

0 = 3.7e^{-0.0714t}\\\\t = \infty

Evaluate the improper integral:

\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\

\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.

\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad

Convert to rev:

51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}

8 0
3 years ago
A 350 gram mass is tied to a string and spin in a horizontal circle with a radius of 11.0cm.The speed of the mass is held consta
SSSSS [86.1K]

Explanation:

a) The angular speed omega is given defined as

omega = #rev/period = 1 rev/(0.6 s) = 1.7 rev/s

= (1.7 rev/s)×(2pi/rev) = 10.5 rad/s

b) v = (omega)×r

= (10.5 rad/s)(0.110 m)

= 1.16 m/s

7 0
3 years ago
A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all fric
Digiron [165]

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

W=K_f -K_i

where

W is the work done

K_f is the final kinetic energy of the cart

K_i is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

W=K_f

But the work is equal to the product between the pushing force F and the displacement, so

W=Fd=(40.0 N)(12.0 m)=480 J

So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is

K_f=\frac{1}{2}mv^2 (1)

where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s



7 0
3 years ago
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