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zysi [14]
3 years ago
8

PLEASE HELP!!

Physics
2 answers:
vesna_86 [32]3 years ago
8 0

It is A. Wind Blow from one location to another.

Blizzard [7]3 years ago
5 0
Wind blows, and water flows, from higher pressure to lower pressure.
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Water exits straight down from a faucet with a 1.96-cm diameter at a speed of 0.55 m/s. The volume flow rate of the water as it
d1i1m1o1n [39]

Answer:

Q = 165.95 cm³ / s,  1)    v = \sqrt{0.55^2 + 19.6 y},  2)  v = 2.05 m / s,

3)  d₂ = 1.014 cm

Explanation:

This is a fluid mechanics exercise

1) the continuity equation is

         Q = v A

where Q is the flow rate, A is area and v is the velocity

         

the area of ​​a circle is

        A = π r²

radius and diameter are related

        r = d / 2

substituting

       A = π d²/4

       Q = π/4   v d²

let's reduce the magnitudes

       v = 0.55 m / s = 55 cm / s

let's calculate

       Q = π/4   55   1.96²

       Q = 165.95 cm³ / s

If we focus on a water particle and apply the zimematics equations

        v² = v₀² + 2 g y

where the initial velocity is v₀ = 0.55 m / s

        v = \sqrt{0.55^2 + 2  \ 9.8\  y}

        v = \sqrt{0.55^2 + 19.6 y}

2) ask to calculate the velocity for y = 0.2 m

        v = \sqrt{0.55^2 + 19.6 \ 0.2}

        v = 2.05 m / s

3) We write the continuous equation for this point 2

        Q = v₂ A₂

        A₂ = Q / v₂

let us reduce to the same units of the SI system

        Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s

        A₂ = 165.95 10⁻⁶ / 2.05

        A₂ = 80,759 10⁻⁶ m²

area is

        A₂ = π/4   d₂²

        d₂ = \sqrt{4  A_2 / \pi }

        d₂ = \sqrt{ \frac{4 \ 80.759 \ 10^{-6} }{\pi } }

        d₂ = 10.14 10⁻³ m

        d₂ = 1.014 cm

4 0
3 years ago
In an experiment, an object is heated.
mariarad [96]

Answer:

B) 300 J/°C

Explanation:

Q = 3kJ = 3000J

ΔT = 10°C

3000J/10°C = 300J/°C

6 0
2 years ago
I)
Natalija [7]

Answer:

I think its (A)The specific heat of an object explains  how easily it changes temperatures.

Explanation:

6 0
2 years ago
Light travels in a straight line at a constant speed of 300000 m/s. What is the acceleration of light?
mafiozo [28]

Answer:

As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.

There is no rate of change of speed, so there is no acceleration.

  • <u>0 m/s²</u> is the right answer.
4 0
3 years ago
A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0
IgorLugansk [536]

Answer:

Magnitude of the force is

F_3 = 2.83

direction of the force is given as

\theta = 45 degree West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

\vec F_1 + \vec F_2 + \vec F_3 = 0

here we know that first force is of magnitude 2 N towards east

\vec F_1 = 2 \hat i N

second force is also of 2.0 N due North

\vec F_2 = 2 \hat j

now from above equation

2\hat i + 2\hat j + \vec F_3 = 0

\vec F_3 = -2\hat i - 2\hat j

so magnitude of the force is given as

F_3 = \sqrt{2^2 + 2^2}

F_3 = 2.83

direction of the force is given as

\theta = tan^{-1}\frac{F_y}{F_x}

\theta = tan^{-1}\frac{-2}{-2}

\theta = 45 degree West of South

3 0
2 years ago
Read 2 more answers
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