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3 years ago

When can a theory be modified if a new type of technology allows for new observations that raise new questions?

Physics

2 answers:

Mumz [18]3 years ago

8 0

Answer: B-After new hypotheses related to the theory are tested in experiments.

Explanation:

A scientific theory can be define as a proposed explanation for a natural event or phenomena that is obtain after repeated experimental procedures. A scientific theory can be changed or amendments can be done in it's content with new discoveries and related evidences.

The implementation of the new technology can reduce the time to conduct the experiments and will also improve the accuracy. Therefore, a new hypothesis can be generated on the basis of the pre-proposed theory by utilizing the new technology.

olga nikolaevna [1]3 years ago

4 0

Answer:

id say B, but not sure

Explanation:

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Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) High and low pressures in kilopascals:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures in pounds per square inch:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures in meter water column in meters water column:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0

3 years ago

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table.

emmainna [20.7K]

C. It doubles I hope this helps and good luck

6 0

3 years ago

Read 2 more answers

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long)

Lesechka [4]

The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?

Answer:

34.6°

Explanation:

To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically

i.e. 29/2 = 14.5

We can solve this through trigonometry.

Let the angle between the ray and the vertical plane mirror is known as α

tan α = 10/14.5

α = $tan^{-1} (10/14.5)$ = 34.6°