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2 years ago

What are some possible reasons for the sudden development of the cell theory

Engineering

1 answer:

11Alexandr11 [23.1K]2 years ago

4 0

Answer:

microscope technology improved. Schleiden made his discovery, that caught the interest of other scientists, who went on to add new discoveries.

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A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of

yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

$V_d=2.5\times 10^{-3} m^3$

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

$P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}$

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0

3 years ago

When buttons or switches are pressed by humans for arbitrary periods of time, we need to convert a signal level to a pulse. In t

ddd [48]

Answer:

The FSM uses the states along with the generation at the P output on each of the positive edges of the CLK. The memory stores the previous state in the machine and the decoder generates a P output based on the previous state.

Explanation:

The code is in the image.

6 0

3 years ago

5. Name two health problems that fume can cause?<br> a)....<br> b)......

Vlad [161]

Answer:

A) Cancer of the Lungs

B)Larynx and Urinary Tract, as well as nervous system and kidney damage

Explanation:

5 0

3 years ago

Read 2 more answers

Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w

I am Lyosha [343]

Answer:

The required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm² and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2 so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm to limit the current to 0.53 A with current density of 459 A/cm² .

6 0

2 years ago

Read 2 more answers

The bar DA is rigid and is originally held in the horizontal position when the weight W is supported from C. The wires are made

Rom4ik [11]

Answer:

W = 112 lb

Explanation:

Given:

- δb = 0.025 in

- E = 29000 ksi (A-36)

- Area A_de = 0.002 in^2

Find:

Compute Weight W attached at C

Solution:

- Use proportion to determine δd:

δd/5 = δb/3

δd = (5/3) * 0.025

δd = 0.0417 in

- Compute εde i.e strain in DE:

εde = δd / Lde

εde = 0.0417 / 3*12

εde = 0.00116

- Compute stress in DE, σde:

σde = E*εde

σde = 29000*0.00116

σde = 33.56 ksi

- Compute the Force F_de:

F_de = σde *A_de

F_de = 33.56*0.002

F_de = 0.0672 kips

- Equilibrium conditions apply:

(M)_a = 0

3*W - 5*F_de = 0

W = (5/3)*F_de

W = (5/3)* 0.0672 = 112 lb

4 0

2 years ago