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Vaselesa [24]
3 years ago
9

before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t

ension in the vehicle's service manual. who is correct?
Engineering
1 answer:
Vsevolod [243]3 years ago
5 0

Answer:

Technician b is correct

Explanation:

Before adjusting drive-belt tension, it is very important to check the vehicle workshop manual for specified belt tension, so that you can match your reading against the specification in the vehicle's service manual. If the tension reading you have matches the suggested reading in the vehicle's service manual and the belt is not damaged then you do not need to proceed any further. But if the reading does not match, then you can adjust the belt tension.

Therefore, technician b is correct.

You might be interested in
List one advantage and one disadvantage of the use of the commutator?
Mariulka [41]

Answer and  Explanation:

Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor

ADVANTAGE OF COMMUTATOR The main advantage of the commutator in DC motor is to keep keep the direction of the toque always in the same direction by changing the current direction

DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.

3 0
3 years ago
A heat exchanger takes compressed liquid water and converts it into steam with a temperature and pressure of 20 Mpa and 480 C; r
belka [17]

Answer:

See explaination

Explanation:

Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.

Please kindly check attachment for the step by step solution of the given problem.

6 0
3 years ago
True or false tensile forces are smaller in arch bridges
Ivan

Answer:

True

Explanation:

The tensile forces are small in most arches and usually negligible.

4 0
2 years ago
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0
3 years ago
Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl
Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

A_1 = 1 m²

P_1 = 100 kPa

V_1 = 180 m/s

Flow rate

Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s

Volumetric flow rate = 180 m³/s

Mass flow rate

\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s

Mass flow rate = 213.4 kg/s

3 0
3 years ago
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