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igomit [66]
3 years ago
9

help please. In 1961 the mlar atomic mass of carbon 12 was defined to be exactly 12g/mol (a value agreed upon by chemists and ph

ysicists) and the molar atomic masses of all the elements were re-evaluated according to this new standard. . This made the re-evaluated molar mass of oxygen-16 become 15.994915g/mol. Avogadro's number is defined as the number of atoms in the molar atomic mass of an element. The present accpted value for Avogadros' number is 6.0221367x10^23/mol. . What was the value o Avogadro's number used by physicists before 1961? .
Physics
1 answer:
Ne4ueva [31]3 years ago
4 0
The first notion of the number of molecules per atom <span>was calculated by Josef Loschmidt which he endeavored to complete in 1895. In his experiment, he determined the number of molecules in one cubic centimeter of gas under STP. this is equal to approx 2.6 x 10^19 molecules. The former is called "Loschmidt's Constant" instead.</span>
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On a Vernier Caliper, how do you know which mark to use on the very top scale?
madreJ [45]

<u>Answer</u>

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use<em>. </em>


<u>Explanation</u>

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale.<em> To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>

4 0
3 years ago
Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?
nikitadnepr [17]

Answer:

The appropriate response is "Optical printer ".

Explanation:

  • A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses.
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4 0
4 years ago
How does the pupillary response prevent injury? What would happen without it?
Aleksandr [31]
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6 0
3 years ago
Read 2 more answers
A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.
alisha [4.7K]
A) 

     It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s
 
     Through Definition of Velocity, comes:

\Delta v=  \frac{\Delta S}{\Delta t}  \\ v_x= \frac{36}{2}  \\ \boxed {v_{x}=18m/s}


B)
 
     Using the Velocity Hourly Equation in vertical direction, we have:

v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}
  
     The angle of impact is given by:

cos(\theta) =\frac{v_{x}}{v_{y}}  \\ cos(\theta) = \frac{18}{20}  \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}


If you notice any mistake in my english, please let me know, because i am not native.

7 0
3 years ago
Read 2 more answers
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

7 0
3 years ago
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