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3 years ago

How are desert plants adapted to their climate?

1 answer:

8 0

They are made for the climate but to answer your question they suck up water and slowly consume it until the next rainfall comes then they suck again

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What is happening in the picture?

Veseljchak [2.6K]

The gravity is pushing rhe boat down

3 0

3 years ago

Read 2 more answers

What is (a) the wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz?

yulyashka [42]

The wavenumber and (b) the wavelength of the radiation used by an fm radio transmitter broadcasting at 92. 0 mhz will be 31.25 * $10^{2}$ $m^{-1}$ and 0.032 * $10^{2}$ m respectively

Forms of electromagnetic radiation like radio waves, light waves or infrared (heat) waves make characteristic patterns as they travel through space. Each wave has a certain shape and length. The distance between peaks (high points) is called wavelength.

Wavenumber, also called wave number, a unit of frequency, often used in atomic, molecular, and nuclear spectroscopy, equal to the true frequency divided by the speed of the wave and thus equal to the number of waves in a unit distance.

wavelength = ?

frequency = 92 m Hz = 92 * $10^{6}$ Hz

speed of light = 3 * $10^{8}$ m/s

speed of light = frequency * wavelength

wavelength = speed of light / frequency

= 3 * $10^{8}$ / 92 * $10^{6}$

= 0.032 * $10^{2}$ m

wavenumber = 1 / wavelength

= 1 / 0.032 * $10^{2}$ m

= 31.25 * $10^{2}$ $m^{-1}$

To learn more about electromagnetic radiation here

brainly.com/question/10759891

#SPJ4

5 0

1 year ago

If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?

Alisiya [41]

Answer:

The spring constant is 3750 N/m

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m

4 0

3 years ago

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

atroni [7]

(a) $2.79 rev/s^2$

The angular acceleration can be calculated by using the following equation:

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha$ is the angular acceleration

$\theta=50.0 rev$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\alpha$ , we find

$\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2$

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 20.0 rev/s$ is the final angular speed

$\omega_i = 11.0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s$

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

$\alpha = \frac{\omega_f-\omega_i}{t}$

where

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

t is the time

Solving for t, we find

$t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s$

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where:

$\omega_f = 11.0 rev/s$ is the final angular speed

$\omega_i = 0 rev/s$ is the initial angular speed

$\alpha=2.79 rev/s^2$ is the angular acceleration

$\theta=?$ is the number of revolutions made by the disk while accelerating

Solving the equation for $\theta$ , we find

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev$

4 0

3 years ago

A 53.3 kg woman slides down a 35.0° hill with an acceleration of 4.10 m/s. What is the friction force acting on the woman?

lorasvet [3.4K]

Answer:

I attached an image that should help.

Explanation:

Check it out.

5 0

3 years ago