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3 years ago

Explain how rock D is the best rock

Physics

1 answer:

Naya [18.7K]3 years ago

7 0

The best rock is D because it is big and heavy out of all of them. Rock A would fling too far because it is the lightest of all rocks as shown.

B and C would probably fling through the target because it is just the right size.

Besides, Depending on their weight of each rock, rock D is heavy because it is big. Rock A is too light to fling or land on the target. This is my hypothesis.

I don't really get how would rock D be the best one, but upon your question, this is all i know.

Thank you for posting your question, hope this helps, have a great day. please mark brainliest so i can answer more of your questions.

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Photovoltaic cells: a) have become more economical to produce and use over the past 25 years. b) are the most efficient means of

Yanka [14]

Photovoltaic cells are the most efficient means of converting solar energy to electricity. Option b is correct.

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2 years ago

A moving fan continues to move for a while even after switched off, why?

dem82 [27]

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3 years ago

Read 2 more answers

before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.

Virty [35]

The momentum of block B after the collision is -50 kg m/s.

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

$p_A + p_B = p'_A + p'_B$

where:

$p_A = -100 kg m/s$ is the momentum of block A before the collision

$p_B = -150 kg m/s$ is the momentum of block B before the collision

$p'_A = -200 kg m/s$ is the momentum of block A after the collision

$p'_B$ is the momentum of block B after the collision

Solving for $p'_B$ , we find:

$p'_B = p_A + p_B - p'_A = -100 +(-150) - (-200)=-50 kg m/s$

So, the momentum of block B after the collision is -50 kg m/s.

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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: $q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}$