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olga_2 [115]
3 years ago
10

Help me with this review question please.

Physics
1 answer:
QveST [7]3 years ago
6 0

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

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The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle
Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

         Re = ρ v D /μ

The units of each term are

       ρ = [kg / m³]

       v = [m / s]

      D = [m]

      μ = [Pa s]

The pressure

      Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

      μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

      Re = [kg / m³] [m / s] [m] / [kg / m s]

      Re = [kg / m s] / [m s / kg]

      RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity       μ = 11.2 10⁻⁶ Pa s

the density  ρ = 0.656 kg / m³

       D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

       Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

       Re = 1.19 10⁴

4 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
The magnitude of the tidal force between the International Space Station (ISS) and a nearby astronaut on a spacewalk is approxim
vovikov84 [41]

Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

      F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

      F = 2 6.67 10⁻¹¹ 135  5.98 10²⁴ 13 / (6.79 10⁶)³

      F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

4 0
3 years ago
Jupiter has the strongest gravity of any planet in our solar system. Which quantity will be a maximum there?
Mamont248 [21]

weight will be maximum there

3 0
2 years ago
This is for physical science. Can someone help me?
Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
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