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Yuliya22 [10]
3 years ago
6

The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind

Physics
1 answer:
denis23 [38]3 years ago
6 0
Meteorites is the answer

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In three sentences, please describe SIMPLE HARMONIC motion, and give two examples. Thank you! :-)
11Alexandr11 [23.1K]
A system that repeats to and from its mean or rest point. that executes harmonic motion. a few examples I've heard of are since the springtime a mass-spring system,a swing, simple pendulum, one more example is a steel ball rolling in a curved is this what you need or do you need three more sentences dish. to get S.H.M a body just displaced away from the resting position and of course then is released. the human body oscillates due to the reinforce that pulls it back do you need anything else answered on this and I'll answer it
3 0
3 years ago
A 1.00-kg mass is attatched to a string 1.0m long and completes a horizontal circle in 0.25s. What is the centripetal accelerati
liraira [26]

-- The string is 1 m long.  That's the radius of the circle that the mass is
traveling in.  The circumference of the circle is  (π) x (2R) = 2π meters .

-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .

-- Centripetal acceleration is  V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²

-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =

                                                         64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .

</span>
That's it.  It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span> 
</span>
If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.

You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping. 
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour !  This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.


3 0
3 years ago
The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of
Alenkasestr [34]

Answer:

a) a=(-2,-6,0)m/s^2, with a magnitude of 6.3m/s^2

b) \frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2, with a magnitude of 0.76Kgm/s^2

c) F=(-0.24,-0.72,0)N, with a magnitude of 0.76N

Explanation:

We have:

v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg

We can calculate each component of the acceleration using its definition a=\frac{\Delta v}{\Delta t}

a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\

The rate of change of momentum of the ball is \frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma

So for each coordinate:

\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N

<em>(N and </em>Kgm/s^2<em> are the same unit).</em>

3 0
3 years ago
Pulsars "blink" because they _____.
rosijanka [135]
A pulsar, or a pulsing star, is a highly magnetized neutron star that emits a beam of electromagnetic radiation. So they blink when they are rotating because the beam of radiation they emit can only be seen when it is facing the Earth.
Hope this helps. 
4 0
3 years ago
What happens to a planes wing when air moves taster over the top of?
Anon25 [30]

Answer:

it begins to decrease it's altitude​

Explanation:

5 0
3 years ago
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