If my mass is 105 kilograms; what is my weight on Earth?

A small 0.13 kg metal ball is tied to a very light (essentially massless) string that is 0.70 m long. The string is attached to

beks73 [17]

Answer:

Explanation:

Let l be th length of pendulum

loss of height

= mg ( l - l cos50)

= mg l ( 1-cos50)

1/2 mv² = mgl ( 1-cos50)

v = √[2gl( 1- cos50)]

= √( 2 x 9.8 x .7 x ( 1-cos50)

= 2.2 m / s

speed at the bottom = 2.2 m /s

b )

centripetal acceleration

= v² / r

= 2.2 x 2.2 / .7

= 6.9 m /s²

C )

If T be the tension

T - mg = mv² / r

T = mg + mv² / r

= .13 X 9.8 + .13 X 6.9

= 2.17 N

7 0

3 years ago

You are parking your vehicle downhill next to a curb. how should you position the wheels

REY [17]

When you stop your car headed downhill, turnyour front wheels toward the curb.

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