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Sindrei [870]
2 years ago
6

What is the scenitific notatiin for 0.000326​

Physics
1 answer:
AleksAgata [21]2 years ago
7 0

Answer:

3.26 x 10^-4

Explanation:

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Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small
Vaselesa [24]

Answer:

The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is

8.00 x 10-13J

Explanation:

In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).

Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)

Although, the kinetic energy is converted to potential energy in Coulomb's law equation.

That is,

1/2(mv^2) = (K* q1q2)/r

Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter

6 0
2 years ago
A scuba diver and her gear displace a volume of 67.0 l and have a total mass of 64.0 kg. (a) what is the buoyant force on the di
slavikrds [6]
Buoyant force is the force that is a result from the pressure exerted by a fluid on the object. We calculate this value by using the Archimedes principle where it says that the upward buoyant force that is being exerted to a body that is immersed in the fluid is equal to the fluid's weight that the object has displaced. Buoyant force always acts opposing the direction of weight. We calculate as follows:

Fb = W
Fb = mass (acceleration due to gravity)
Fb = 64.0 kg ( 9.81 m/s^2)
Fb = 627.84 kg m/s^2

Therefore, the buoyant force that is exerted on the diver in the sea water would be 627.84 N
4 0
3 years ago
How do the principles of convection, conduction, and radiation explain how the water in the saucepan gets hot?
RSB [31]
<span>Heat comes from stove flame to the sauce pan by radiation through infrared energy, heat conducts the metal of the sauce pan; Convection brings cool water to the hot surface at the bottom of the hot sauce pan until all or most of the water is hot enough to boil.</span>
3 0
3 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
Explain nanotechnology. Its advantages, disadvantages with examples or application? ​
Volgvan

Answer:

PURPOSE

This information bulletin clarifies the Department of Building and Safety’s (LADBS) requirement

relating to the equipment (product) and wiring approval as delineated in Sections 110.2, 110.3,

93.0401, 93.0402, and 93.0403 of the 2017 edition of the City of Los Angeles Electrical Code (LAEC).

DEFINITIONS

Approved - Acceptable to authority having jurisdiction. {California Electrical Code (CEC), Article 100}

Authority Having Jurisdiction – An organization, office, or individual responsible (i.e., LADBS

Electrical Testing Laboratory or Inspection) for enforcing the requirements of a code or standard, or

for approving equipment, material, an installation, or a procedure. (CEC, Article 100)

Equipment - A general term, including material, fittings, devices, appliances, luminaires, apparatus,

machinery, and the like used as part of, or in connection with, an electrical installation. (CEC, Article

100)

Identified - Recognizable as suitable for the specific purpose, function, use, environment, application,

and so forth, where described in particular code requirement. (CEC, Article 100)

Labeled - Equipment or material to which has been attached a label, symbol, or other accepted

identifying mark (i.e., embossed laboratory logo) of a recognized (approved) testing agency (see

definition) and concerned with product evaluation, that maintains periodic inspection of production of

labeled equipment or materials, and by whose labeling the manufacturer indicates compliance with

recognized national safety standards (see definition). (CEC, Article 100)

7 0
3 years ago
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