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3 years ago

6

What happens when the data in an experiment does not support the hypothesis?

1 answer:

liq [111]3 years ago

6 0

Say in your conclusion that your Hypothesis was incorrect

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Sunlight travels 150,000,000 km from the sun to the earth because of?

padilas [110]

The Speed of Light.
Photons emitted from the surface of the sun to travel across the vacuum of space to reach out eyes

8 0

3 years ago

g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell

olga55 [171]

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

The current density is $J = 4*10^6 \ A/m^2$

The conductivity is $\sigma = 2*10^{7} \ S/m$

Generally conductivity is mathematically represented as

$\sigma = \frac{l}{RA}$

Here R is the resistance which is mathematically represented as

$R = \frac{V}{I}$

Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

=> $\sigma = \frac{l}{\frac{V}{J}}$

=> $V = \frac{l * J}{\sigma }$

=> $V = \frac{10 * 4*10^6}{2*10^{7} }$

=> $V = 2 V$

5 0

2 years ago

the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert

AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure $P$ is the force $F$ exerted by a gas, a liquid or a solid on a surface (or area) $A$ , its unit is Pascal $Pa$ which is equal to $N/m^{2}$ and its formula is:

$P=\frac{F}{A}$ (1)

In this case we have the surface of a sole of a tennis shoe:

$A=0.0290 m^{2}$ (2)

And the mass $m$ of the person who wears it:

$m=65 kg$

On the other hand, we know the weight is the force $F$ the Earth exerts on people and objects due gravity $g$ :

$F=m.g=(65 kg)(9.8m/^{2})$

$F=637N$ (3)

Substituting (2) and (3) in (1):

$P=\frac{637N}{0.0290 m^{2}}$ (4)

Finally:

$P=21965.517 Pa$ This is the pressure the shoe exert on the ground

5 0

2 years ago

Mass is a fundamental quantity

Oliga [24]

Explanation:

okk hope you like then comment

5 0

3 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

or

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

6 0

3 years ago