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3 years ago

Calculate the potential required to push eleven amperes through a 20 ohm resistance

Physics

2 answers:

den301095 [7]3 years ago

5 0

Answer:

The answer is 220

Explanation:

Volts = current × resistance

Current = 11

Resistance = 20

Volts = 11 × 20

= 220

likoan [24]3 years ago

3 0

<h2>The potential required = 220 V</h2>

Explanation:

Here the current to flow I = 11 A

and the resistance of conductor R = 20 ohm

The potential difference = I R

= 11 x 20 = 220 V

Thus 220 volt of potential is required

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After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress strain curve and calcul

Fittoniya [83]

Answer:

Part a: <em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

Part b: <em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Part c: <em>The value of Young's modulus at given point is 172 GPa.</em>

Part d: <em>The percentage elongation is 18.55%.</em>

Part e: The percentage reduction in area is 15.81%

Explanation:

From the complete question the data is provided for various Loads in ductile testing machine for a sample of d0=20 mm and l0=40mm. The plot is drawn between stress and strain whose values are calculated using following formulae. The corresponding values are attached with the solution.

The engineering-stress is given as

$\sigma=\frac{F}{A}\\\sigma=\frac{F}{\pi \frac{d_0^2}{4}}\\\sigma=\frac{F}{\pi \frac{(20 \times 10^-3)^2}{4}}\\\sigma=\frac{F}{3.14 \times 10^{-4}}$

Here F are different values of the load

Now Strain is given as

$\epsilon=\frac{l-l_0}{l_0}\\\epsilon=\frac{\Delta l}{40}\\$

So the curve is plotted and is attached.

<em>The value of yield strength at 0.2% offset is obtained from the engineering stress-strain curve which is given as 274 MPa.</em>

<em>The value of tensile strength is obtained from the engineering stress-strain curve which is given as 417 MPa</em>

Young's Modulus is given as

$E=\frac{\sigma}{\epsilon}\\E=\frac{238 /times 10^6}{0.00138}\\E=172,000 MPa\\E=172 GPa$

<em>The value of Young's modulus at given point is 172 GPa.</em>

The percentage elongation is given as

$Elongation=\frac{l_f-l_0}{l_0} \times 100\\Elongation=\frac{47.42-40}{40}\times 100\\Elongation=18.55 \%\\$

So the percentage elongation is 18.55%

The reduction in area is given as

$Reduction=\frac{A_0-A_n}{A_0} \times 100\\Reduction=\frac{\pi \frac{d_0^2}{4}-\pi \frac{d_n^2}{4}}{\pi \frac{d_0^2}{4}}\times 100\\Reduction=\frac{{d_0^2}-{d_n^2}}{{d_0^2}} \times 100\\Reduction=\frac{{20^2}-{18.35^2}}{{20^2}} \times 100\\Reduction=15.81\%$

So the reduction in area is 15.81%

6 0

3 years ago

(Subject is Astronomy, but that's not an option.)

jasenka [17]

Answer:

Astrometry (D)

Explanation:

4 0

3 years ago

Read 2 more answers

Suppose a motor connected to a 120 V source draws 20.0 A when it first starts.

lisabon 2012 [21]

Answer:

Explanation:

8 0

3 years ago

Which option best describes the increase in pressure when the volume of a plastic bottle decreases?

Allisa [31]

Answer:

B) uniform

Explanation:

We can solve this problem by using Boyle's law, which states that:

<em>For a constant mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume.</em>

Mathematically:

where

p is the pressure of the gas

V is its volume

If we apply the equation to the bottle in the problem, we see that:

- when the volume of the bottle (and therefore, of the gas inside) decreases, than the pressure will increase

- viceversa, when the volume of the bottle increases, the pressure will decrease

The amount by which the pressure increases is inversely proportional to the decrease in volume, so the answer depends on how the volume of the bottle decreases: however, if the volume of the bottle decreases uniformly, then we can say that the pressure inside the bottle will also increases uniformly.

MARK THIS ANSWER AS THE BRAINLIEST PLEASE ❤️

6 0

3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

4 years ago