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1 year ago

10

What is the potiental energy of a 3 kg ball that is on the ground

1 answer:

Llana [10]1 year ago

8 0

Answer:

147.15

Explanation:

147.15 is the answer

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Which of the following is not a primary or fundamental dimension? (a)-mass m (b)-length L (c)- timer t (d)-volume V

jolli1 [7]

Answer:

Volume is not the fundamental dimension. So, option d is correct.

Explanation:

Step1

Fundamental dimension is the dimension in which other quantities depend. These are the basic dimensions. Three basic fundaments dimensions are mass, length and time that is represented as MLT respectively.

Step2

Volume is not the fundamental dimension among them as the volume is cube of length dimension. Thus, volume depends upon length.

Thus, Volume is not the fundamental dimension. So, option d is correct.

8 0

3 years ago

As you have learned, not all motor oils are the same. What are two things that make them different?.

saul85 [17]

Answer:

quality ingredients and exceeding industry standards

Explanation:

5 0

1 year ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

2 years ago

An isothermal CSTR with a first order irreversible reaction A â> B (and rA[mol/(ft3*min)] = - 0.5 CA) has a constant flow rat

worty [1.4K]

Can you help me by solving this question

7 0

2 years ago

Meaning of temporary stitches

algol13

Answer:

They are use to hold garment or fabric pieces together before pernament stitches are made

6 0

2 years ago

Read 2 more answers