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3 years ago

Power is a fundamental dimension. a) True b) False

Engineering

1 answer:

Ronch [10]3 years ago

4 0

Answer:

The statement is false.

Explanation:

Fundamental dimensions are those dimensions that cannot be derived from any combination of physical quantities in the universe and these quantities act as base quantities from which all other physical quantities are derived.

The fundamental dimensions in the nature are:

1) Length (L)

2) Mass (M)

3) Time (T)

4) Current (A)

5) Temperature (K)

6) Amount/ Quantity (MOLE)

7) Luminous intensity (CANDELA)

All the other physical quantities are derived from a combination of these fundamental dimensions.

Power can be represented as $\frac{ML^{2}}{T^{3}}$ and hence it is a derived quantity.

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Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF

Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

4 0

3 years ago

The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu

marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0

3 years ago

Water flows at a rate of 10 gallons per minute in a new horizontal 0.75?in. diameter galvanized iron pipe. Determine the pressur

ruslelena [56]

Answer:

$\frac{\delta p }{l} = 30.4 lb/ft^3$

Explanation:

Given data:

flow rate = 10 gallon per minute = 0.0223 ft^3/sec

diameter = 0.75 inch

we know discharge is given as

Q = VA

solve for velocity V = \frac{Q}{A}[/tex]

$V = \frac{0.223}{\frac{\pi}{4} \frac{0.75}{12}}$

V = 7.27 ft/sec

we know that Reynold number

$Re = \frac{VD}{\nu}$

$Re = \frac{7.27 \times \frac{0.75}{12}}{1.21\times 10^{-5}}$

$Re = 3.76 \times 10^4$

calculate the $\frac{\epsilon }{D}$ ratio to determine the fanning friction f

$\frac{\epsilon }{D} = \frac{0.0005}{\frac{0.75}{12}} = 0.008$

from moody diagram f value corresonding to Re and $\frac{\epsilon }{D}$ is 0.037

for horizontal pipe

$\delta p = \frac{f l \rho v^2}{2D}$

$\frac{\delta p }{l} = \frac{1 \times 0.037 \times 1.94 \times 7.27}{\frac{0.75}{12}}$

where 1.94 slug/ft^3is density of water

$\frac{\delta p }{l} = 30.4 lb/ft^3$

3 0

3 years ago

1. Describe simply what will happen to an airplane in flight in the following conditions:

notka56 [123]

Answer:

For the two you haven't answered: (Drag greater than thrust, lift greater than weight) It will accelerate backwards (decelerate) and upwards

(Lift greater than weight, thrust greater than drag) accelerate upwards and forwards.

4 0

3 years ago

1. Under what conditions can soils be chemically stabilized?

marshall27 [118]

Answer:

All will be Explained below.

Explanation:

1) Under which Condition can a soil be chemically Stabilize.

Answer

a). Plasticity Index :A soil with a high value of plasticity Index is not good for various engineering projects. The introduction of line helps in reducing plasticity due cation exchange reaction.Pozzolanic reaction over time reduces plasticity and increase index strength due to the formation of calcium - silicate hydrate.

7 0

3 years ago