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Firdavs [7]
3 years ago
12

A current of 0.3 ampere flows through a 2-ohm resistor. At what rate does the resistor release heat?

Engineering
1 answer:
beks73 [17]3 years ago
6 0

Answer:

D. 018 W

Explanation:

Power = current × voltage

P = IV

From Ohm's law, voltage = current × resistance

V = IR

Therefore:

P = I²R

Given I = 0.3 A and R = 2 Ω:

P = (0.3 A)² (2 Ω)

P = 0.18 W

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Water flows through a pipe and enters a section where the cross sectional area is larger. Viscosity, friction, and gravitational
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(A) and (D)

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3 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

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3 years ago
What are the benefits of using the engineering design process? combining scientific knowledge and creativity in a rigid and stru
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combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving

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