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Gre4nikov [31]
4 years ago
11

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

ght is 500 N. Rotational speed is 1774. How many kW of power will be developed?
Engineering
1 answer:
miss Akunina [59]4 years ago
4 0

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

P = T\times \omega

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

\omega angular speed =\frac{2 \pi N}{60}

Therefore Power is

P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt

P = 80.922 KW

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Lisa [10]

Answer:

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Explanation:

7 0
2 years ago
For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What
Alenkinab [10]

Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

k = E * A / l

k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

5 0
3 years ago
A ________ is a condition or capability needed by a user to solve a problem or achieve an objective that satisfies a standard or
Aliun [14]
A successful brain to help
7 0
3 years ago
You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an
Ymorist [56]

Answer:

t'_{1\2} = 6.6 sec

Explanation:

the half life of the given circuit is given by

t_{1\2} =\tau ln2

where [/tex]\tau = RC[/tex]

t_{1\2} = RCln2

Given t_{1\2} = 3 sec

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

t'_{1\2} =R'Cln2

Divide equation 2 by 1

\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}

t'_{1\2} = t'_{1\2}\frac{R'}{R}

putting all value we get new half life

t'_{1\2} = 3 * \frac{88}{40}  = 6.6 sec

t'_{1\2} = 6.6 sec

7 0
4 years ago
Please calculate the current for the circuit below
Aleks04 [339]

Answer:

The answer is "I = 0.0085106383 \ A"

Explanation:

Given:

R= 470  \ \Omega \\\\V= 4 \ v

Formula:

\to V=IR\\\\\to I = \frac{V}{R}\\\\

      = \frac{4}{470}\\\\ = 0.0085106383 \ A

3 0
3 years ago
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