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posledela
3 years ago
8

What is a particulate ? Name a couple of examples.

Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles

Explanation:

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I need an answer asap
gulaghasi [49]
He has a mass of 56 kg.

The equation given is PE = mgh.

PE = 4620 J

h = 8.4

g = 9.8

Therefore:

4620 = 82.32m

m = 4620/82.32
m = 56 (rounded to two significant digits)
5 0
3 years ago
Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds th
Oksi-84 [34.3K]

Answer:

The  frequencies are  f_n  =  n (0.875 )

Explanation:

From the question we are told that

   The speed of the wave is  v  =  0.700 \  m/s

   The  length of vibrating  clothesline is  L  =  40.0 \  cm = 0.4 \ m

Generally the fundamental frequency is  mathematically represented as

        f =  \frac{v}{2 L  }

=>     f =  \frac{ 0.700 }{2 *  0.4   }

=>     f =  0.875 \  Hz

Now  this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of  the fundamental frequency

So  

   The  frequencies are mathematically represented as

       f_n  =  n  * f

=>     f_n  =  n (0.875 )

Where  n  =  1, 2, 3 ....

       

3 0
3 years ago
What is Stefan's botmann's constant​
olga2289 [7]

Answer:

The Stefan–Boltzmann constant (also Stefan's constant), a physical constant denoted by the Greek letter σ (sigma), is the constant of proportionality in the Stefan–Boltzmann law: "the total intensity radiated over all wavelengths increases as the temperature increases", of a black body which is proportional to the ...

4 0
3 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
What is the acceleration of a 1000kg car subject to a 550N net force?
igomit [66]

Answer:

a=550÷1000

a=0.55m/s²

5 0
3 years ago
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