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Lina20 [59]
3 years ago
14

For the following gear train, if the blue gear is moving at 50 rpm, what are the speeds of the other gears?

Engineering
1 answer:
Flauer [41]3 years ago
5 0

Answer:

6

Explanation:

6 teddy bears

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Complex poles cmd zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the listed open-loop mmsfer fuoc
Roman55 [17]

Answer: e is the best answer

Explanation:

8 0
3 years ago
Determine the number of flipflops required to build a binary counter that count from 0 to 2043
Pie

Answer:

10 flip -flops are required to build a binary counter circuit to count to from 0 to 1023 .

Explanation:

3 0
3 years ago
At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide
Xelga [282]

Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0    (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s  (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If  2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0    (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

⇒ vB = 70 mm/s

d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

4 0
3 years ago
Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan
Digiron [165]

Answer:

The final temperature of water is 381.39  °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

Q=mC_p\Delta T

We know that for water

C_p=4.187\ \frac{KJ}{kg.K}

Lets take final temperature of water is T

So

Q=mC_p\Delta T

2960=5\times 4.187(T-240)

T=381.39  °C

So the final temperature of water is 381.39  °C.

7 0
3 years ago
The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?
aalyn [17]

Answer:

891.027 lbm/ft³

Explanation:

See it in the pic

5 0
3 years ago
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