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3 years ago

The voltage across a device and the current through it are:

Engineering

2 answers:

Strike441 [17]3 years ago

7 0

Answer: Charge, Q= 22.131 columbs

Power, P = 196.65 Watts

Explanation:

I(t) = 10( 1 - e(-0.5t)) A

I = dQ/dt

dQ = 10( 1 - e(-0.5t))dt

Integrating we have

Q = 10t - (20/(-0.5))e(-0.5t)

taking the value of t at 1s

we have,

Q = 10 + 20e(-0.5)

Q = 10 + 20/1.6487

Q = 10 + 12.131

Q = 22.131 Columbs

(b) Power

Power = voltage x current

Power = 5cos(2t) x 10(1 - e(-0.5t))

Power at t= 1, we substitute t=1

Power = 5cos(2) x 10(1 - e(-0.5))

Power = 5(0.9995) x 10(1 - 0.60654)

Power = 39.35 x 4.9975 = 196.65 Watts

Power = 196.65 Watts

butalik [34]3 years ago

3 0

Answer:

attached below

Explanation:

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The four applicants for a $40,000 loan are Robin Smith, Amari Jones,

Lapatulllka [165]

Answer:

Clearly its Robin smith

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3 years ago

From the article "Time Travel Is A Fun Science Fiction Story But Could It Be Real?", Albert Einstein's Theory of Relativity is d

zhannawk [14.2K]

Answer:

Option A

Explanation:

As per the article "Time Travel Is A Fun Science Fiction Story But Could It Be Real?" the gravitational field is a representation of curving space and time. As the gravity becomes strong, the space-time get more curved and hence the time gets slower.

Hence, option A is correct

8 0

3 years ago

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

A rigid vessel with a volume of 10 m3 contains a water-vapor mixture at 400 kPa. If the quality is 60 percent, find the mass. Th

trapecia [35]

A) The pressure (p)=400kPa=0.400MPa
Mass of mixture =M
quality (X)= 0.60
Volume of mixture (V)=10 m3
From steam table at P=0.400MPa
Specific volume of saturated water (vf)=0.00108355 m3/kg
Specific volume of saturated steam (vg)=0.46238 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.00108355+0.6(0.46238-0.00108355)]
M=36.04748 kg
If pressure is lowered to 300kPa
p=0.300MPa
From steam table we get,
Specific volume of saturated water (vf)=0.00107317 m3/kg
Specific volume of saturated steam (vg)=0.60576 m3/kg
Therefore, the volume of steam having X=0.6 is given by
V=M[vf+X(vg-vf)]
10= M[0.001007317+0.6(0.60576-0.00107317)]
M= 26.4825 kg

7 0

3 years ago

A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici

Alika [10]

Answer:

The maximum length of a surface flaw is 8.24 μm

8.24 μm

Explanation:

Given that:

The modulus of elasticity E = 69 GPa

The specific surface energy $\delta_s$ = 0.3 J/m²