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At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing

Inessa [10]

Answer:

vB = - 0.176 m/s (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)

⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)

⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒ vB = - 0.176 m/s (↓-)

The pic can show the question.

7 0

3 years ago

Read 2 more answers

Write a function call with arguments tensPlace, onesPlace, and userInt. Be sure to pass the first two arguments as pointers. Sam

Trava [24]

Answer:

#include <stdio.h>

void SplitIntoTensOnes(int* tensDigit, int* onesDigit, int DecVal){

*tensDigit = (DecVal / 10) % 10;

*onesDigit = DecVal % 10;

return;

}

int main(void) {

int tensPlace = 0;

int onesPlace = 0;

int userInt = 0;

userInt = 41;

SplitIntoTensOnes(&tensPlace, &onesPlace, userInt);

printf("tensPlace = %d, onesPlace = %d\n", tensPlace, onesPlace);

return 0;

}

4 0

3 years ago

an inclined manometer is connected to a pitor tube to measure the velocity at the center of a circular duct. If the inclined man

jekas [21]

57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though

8 0

3 years ago

5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h

frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0

3 years ago

Technician A says that as long as the dog (clutch) teeth of the synchronizer are rounded, the synchronizer can be reused. Techni

Sidana [21]

Neither of the technicians are correct.

<h3>Who is a technician?</h3>

A technician simply means an individual who looks after the technical equipments in a company.

In this case, the teeth of the synchronizer are not rounded, the synchronizer can be reused. This shows that he is incorrect likewise technician B.

Learn more about technicians on:

brainly.com/question/1548867

#SPJ12

6 0

2 years ago