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2 years ago

8

A tiger cub has a pattern of stripes on it for that is similar to that of his parents where are the instructions stored that pro

vide information for a tigers for a pattern

1 answer:

ludmilkaskok [199]2 years ago

3 0

probably in it's chromosomes

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Single point cutting tool removes material from a rotating work piece to generate a cylinder is called • Facing Tuming • Both 1

madam [21]

Answer:Turning

Explanation: Turning is the process in which the work piece is subjected to machining so that excess part is removed with the help of rotation by turning machine or lathe machine.The cutter tool is used for cutting the excess of the work piece and it is mostly single-pointed so that give accurate removal of the excess of work piece.At times , according to the requirement multi-pointed tool is also used Therefore, the correct option is turning.

6 0

3 years ago

An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric

lawyer [7]

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

$h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}$

At temperature= 50°C

$h_2=73 \frac{KJ}{kg}$

$So\ the\ heater\ load =h_2-h_1$

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

3 0

3 years ago

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago

A quantity of water within a piston-cylinder assembly executes a Carnot power cycle. During isothermal expansion, the water is h

Harman [31]

Answer:

Answer with Explanation is in the following attachments.

Explanation:

7 0

3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

2 years ago