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3 years ago

Is it more fun on a swing with low or high amplitude? Explain.

1 answer:

7 0

A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity. In the example of the playground swing, the swing is supported by chains that are attached to fixed points at the top of the swing set. When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it.

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3. How does the wavelength of green different from violet-indigo?

grandymaker [24]

Answer:

As the full spectrum of visible light travels through a prism, the wavelengths separate into the colors of the rainbow because each color is a different wavelength. Violet has the shortest wavelength, at around 380 nanometers, and red has the longest wavelength, at around 700 nanometers.

Explanation:

6 0

3 years ago

An object has 90,000 J of kinetic energy and is moving at 12 m/s. What is the objects mass?

trapecia [35]

Answer:

plug it into your calculator

Explanation:

Kinetic energy = KE = Joules = J

KE= 1/2mv^2

90,000 = 1/2 m (12^2)

180,000 = m (144)

m = 180,000 / 144

4 0

4 years ago

Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn

RSB [31]

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128 kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

5 0

3 years ago

What effect does observing a substance's physical properties have on the substance?

Free_Kalibri [48]

If you're careful, you ought to be able to observe ANY of these properties
without any effect on the substance:

Absorption, albedo, angular momentum, area, color, concentration,
density, elasticity, electric charge, electrical conductivity, flow rate,
electrical impedance, electric potential, fluidity, length, location, mass,
luminance, luminescence, luster, magnetic field, momentum, opacity,
permeability, permittivity, plasticity, pressure, radiance, solubility, spin,
specific heat, resistivity, reflectivity, refractive index, temperature,
thermal conductivity, velocity, viscosity, or volume.

6 0

3 years ago

Saturated ethylene glycol at 1 atm is heated by a horizontal chromiumplated surface which has a diameter of 200 mm and is mainta

Paha777 [63]

Here is the full question.

Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.

At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.

Estimate the heating power requirement and the rate of evaporation

What fraction is the power requirement of the maximum power associated with the critical heat flux

Answer:

The heating power requirement = 559.2 W

The rate of evaporation = $6.89*10^{-4}kg/s$

The fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

Explanation:

From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and $T_{sat}$ = 470 K for the saturated ethylene glycol.

Value for enthalpy of formation $h_{fg}$ = 812 kJ/kg

Density of saturated ethylene glycol $\rho___l$ = 1111 kg/m³

Surface tension $\sigma = 32.7*10^{-3}N/m$

The heat flux can be calculated by using the formula:

$q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2} [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3$

$= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]$

= 308.56 × 576.6 × 0.1

= 1.78 × 10⁴ W/m²

Now; to find the heating power requirement; we have:

$q_{boil} = q__s }*A S$

= $1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2$

Thus, the heating power requirement = 559.2 W

The rate of evaporation is given as:

$m= \frac{q_{boil}}{h_[fg}}$

= $\frac{559.2}{812*10^3}$

= $6.89*10^{-4}kg/s$

Thus, the rate of evaporation = $6.89*10^{-4}kg/s$

To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.

So, the calculation for the critical heat is given as: $q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}$

= $q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}$

= 200840.08 × 3.37

= 6.77 × 10⁵ W/m²

Finally, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is as follows:

= $\frac{q''s}{q''max}$

= $\frac{1.78*10^4}{6.77*10^5}$

= 0.026

Thus, the fraction of the power requirement of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

7 0

3 years ago