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3 years ago

14

what kind of weather is associated with high pressure?How does density,humidity, and air motion compare to that low pressure sys

tem?

1 answer:

frutty [35]3 years ago

6 0

Usually nice weather , i dont know the answer to the second part

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Which disease is caused by the loss of bone calcium?

Lelechka [254]

rickets is right answer

3 0

3 years ago

Read 2 more answers

Find the speed of a wave with a frequency of 18 Hz and a wavelength of 6 meters. Show work. WILL MARK BRAINLIEST IF CORRECT

Katyanochek1 [597]

Answer:

so i would say 11.4 i dont have work only this link

Explanation:

https://flexbooks.ck12.org/cbook/ck-12-physics-flexbook-2.0/section/11.4/primary/lesson/wave-speed-ms-ps

3 0

3 years ago

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w

garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

$E_s=\dfrac{1}{2}kx^2$

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

$E_P=mgx$

Equating these two energies,

$\dfrac{1}{2}kx^2=mgx$

$\dfrac{k}{m}=\dfrac{2g}{x}$ .............(1)

The expression for the oscillation frequency is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}$ (from equation (1))

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}$

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0

4 years ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

6 0

3 years ago

A bolt is dropped from a bridge under construction, falling 97 m to the valley below the bridge. (a) how much time does it take

exis [7]

The first thing we have to do for this case is write the kinematic equationsto
vf = a * t + vo
rf = a * (t ^ 2/2) + vo * t + ro
Then, for the bolt we have:
100% of your fall:
97 = g * (t ^ 2/2)
clearing t
t = root (2 * ((97) / (9.8)))
t = 4.449260429
89% of your fall:
0.89*97 = g * (t ^ 2/2)
clearing t
t = root (2 * ((0.89 * 97) / (9.8)))
t = 4.197423894
11% of your fall
t = 4.449260429-4.197423894
t = 0.252

To know the speed when the last 11% of your fall begins, you must first know how long it took you to get there:
86.33 = g * (t ^ 2/2)
Determining t:
t = root (2 * ((86.33) / (9.8))) = <span> 4.19742389 </span>s
Then, your speed will be:
vf = (9.8) * (4.19742389) = 41.135 m / s

Speed just before reaching the ground:
The time will be:
t = 0.252 + <span> 4.197423894</span> = <span> 4.449423894</span> s
The speed is
vf = (9.8) * (4.449423894) =<span> <span>43.603</span></span> m / s

answer
(a) t = 0.252 s
(b) 41,135 m / s
(c) 43.603 m / s

6 0

3 years ago