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Simora [160]
1 year ago
12

As a conductor diameter is changed, the circular mil area will change. if a conductor diameter is changed so that the circular m

ils double, the conductor resistance will ________.
Physics
1 answer:
kolezko [41]1 year ago
7 0
The answer would be halve
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A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a
ss7ja [257]

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

w = m*g

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

w = 60*9.81\\w = 588.6[N]

And the pressure:

P=588.6/0.0004\\P=1471500 [Pa]

Because 1 [Pa] = 1 [N/m²]

8 0
3 years ago
What happens when the data in an investigation does not support the origanal hypothesis
alexgriva [62]

Answer: The scientist gives up and starts an investigation on a new topic.

Explanation:

The data is altered so that it supports the original hypothesized. The data is then altered so that it supports the original hypothesis.

5 0
3 years ago
What is the resistance at 20°C of a 2.0-meter length of tungsten wire with a cross-sectional area of 7.9 10^-7
Bad White [126]

Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

8 0
3 years ago
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll
Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

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3 years ago
A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming
worty [1.4K]

d. 49.0 m/s

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3 0
3 years ago
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