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Temka [501]
3 years ago
8

How can humidity be used to predict if it is going to rain?

Physics
1 answer:
postnew [5]3 years ago
3 0
1. Humidity cannot be used to predict rain.
2. I'm pretty sure it's weather but I'm not 100% sure. Maybe like 89% sure.

3. Tempurature doesn't affect humidity.

4. Not sure but I think its the 3rd one
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Distance from any point on a wave to an identical point on the next wave?
lesantik [10]
It's called the "Wavelength". It corresponds to <span>the distance from any point on a wave to an identical point on the next wave and could also be from crest to crest or trough to trough.

Hope this helps !

Photon</span>
6 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
If an object travels at a constant speed in a circular path, the acceleration of the object is:
Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

a = \frac{v^2}{r}

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 0
3 years ago
9. A plane starts at rest &amp; accelerates along the ground before takeoff. It
Phoenix [80]

Answer:

  9.877 m/s^2

Explanation:

The acceleration can be computed from ...

  d = (1/2)at^2

  (1600 m) = (1/2)a(18 s)^2

  a = (1600/162) m/s^2 ≈ 9.877 m/s^2

6 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

6 0
3 years ago
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