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2 years ago

15

Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t

he workpiece and 75 feet back to the welding machine.

1 answer:

Anni [7]2 years ago

8 0

Answer:

AC or

Explanation:

BC

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A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

$emf= N\frac{d \phi}{dt}$

where;

ΔФ is change in magnetic flux

ΔФ = BA = B x πr²

ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

$emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V$

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0

2 years ago

It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of

monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

M_c = F_a*( 42 / 150 ) *88

M_c = 2.5*( 42 / 150 ) *88

M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0

3 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

2 years ago

What is polarized electrical receptacle used for

lidiya [134]

Answer:

they are used for electrical currents so that they can flow along the appropriate wires in the circuit

Explanation:

4 0

3 years ago

Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

Vinil7 [7]

Answer:

1) $V_o = 10 liters$

2) $V_o = 12.26 liters$

Explanation:

For isothermal process n =1

$V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}$

$V_o = 10 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

$actual \ volume = c1\times 10 = 10.3 liters$

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

$V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}$

$V_o = 12.26 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

$actual \volume = c1\times 10 = 11.5 liters$

8 0

3 years ago