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3 years ago

What is the SPEED of a ball that travels 120 m in 6s?

Physics

1 answer:

AVprozaik [17]3 years ago

7 0

Answer:

20m/second

Explanation:

The reason the answer is 20m/second is because to find the speed of the ball in this question you have to divide the distance over the time giving you the result of 20m/second

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A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9

ElenaW [278]

Answer:

The convection coefficient is $15456.48\ W/m^{2}K$

Solution:

Mass flow rate, $\dot{m} = 0.4\ kg$

Inner diameter of the tube, d = 0.014 m

Fluid density, $\rho_{f} = 990\ kg/m^{3}$

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, $P_{r} = 8.6$

Heat flux, $\dot{q} = 71,297\ W/m^{2}$

Viscosity, $\mu = 0.00079\ Ns/m^{2}$

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

$A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}$

Determine the velocity of the fluid inside the tube by mass flow rate:

$\dot{m} = \rho_{f}A_{c}v$

$0.4 = 990\times 1.54\time 10^{- 4}v$

v = 2.624 m/s

Determine the Reynold's Number, $R_{e}$ :

$R_{e} = \frac{\rho_{f}dv}{\mu}$

$R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253$

Thus it is clear that $R_{e}$ > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

$N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}$

$N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42$

Also,

$N_{u} = \frac{dh}{K}$

where

h = convection coefficient

Now,

$292.42 = \frac{0.014\times h}{0.74}$

$h = 15456.48\ W/m^{2}K$

7 0

3 years ago

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used

Blizzard [7]

Answer:

a,b) #_ {electron} = 1.64 10¹⁹ electrons, c) R = 19.54 Ω, d) V = 10.3 V

Explanation:

a and b) The current is defined as the number of electrons that pass per unit of time

let's look for the load

Q = I t

Q = 0.526 5

Q = 2.63 C

Let's use a direct rule of three proportions. If an electron has a charge of 1.6 10⁻¹⁹ C, how many electrons does 2.63 C have?

#_ {electron} = 2.63 C (1 electron / 1.6 10⁻¹⁹)

#_ {electron} = 1.64 10¹⁹ electrons

c) the resistance of a wire is given by

R = ρ l / A

where the resistivity of tungsten is 5.6 10⁻⁸ Ω

the area of the wire is

A = π r2 = π d²/4

we substitute

R = $\rho \ l \ \frac{4}{\pi d^2}$

let's calculate

R = 5.6 10⁻⁸ 0.580 $\frac{4}{ \pi (0.046 \ 10^{-3})^2 }$

R = 19.54 Ω

d) let's use ohm's law

V = i R

V = 0.526 19.54

V = 10.3 V

7 0

3 years ago

What is the wavelength (in air) of a 13.0 khz vlf radio wave?

bonufazy [111]

Since,
Speed = Frequency * WaveLength

=> WaveLength = Speed / Frequency --- (A)

Frequency = 13.0 kHz.
As the radio waves are electromagnetic waves, their speed is equals to the speed of light. Therefore,
Speed = C = $3 * 10^{8} m/ s^{2}$

Plug in the values in equation(A):

A => WaveLength = $\frac{3 * 10^{8}}{13*10^{3}}$

Ans: Wavelength = 23.077 kilometers.
-i

5 0

3 years ago

If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc

prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0

2 years ago

I think you said everything but the answer.

jeka94

Answer:

What does that even mean?

Explanation:

7 0

3 years ago

Read 2 more answers