All categories

3 years ago

Using the data provided below, calculate the corrected wavelength for a spectroscope reading of 6.32.

Physics

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

646.6 nm

Explanation:

Using the data given, we fit a linear model :

Spectroscope Readings Known Helium Wavelengths (nm)

3.65 388.9

4.75 486.6

4.90 501.6

5.65 587.6

6.45 667.8

7.00 706.5

Using technology like excel to fit the model, the regression model obtained is :

Y = 97.96582X + 27.48458

Where, y is the predicted or corrected wavelength value.

This mod could be used to calculate the corrected wavelength for a given Spectroscope value :

Given a Spectroscope value of 6.32 ; the corrected wavelength is obtained by replacing x in the equation by 6.32 and calculate y ;

Y = 97.96582X + 27.48458

x = 6.32

Y = 97.96582(6.32) + 27.48458

Y = 619.1439824 + 27.48458

Y = 646.6285624

Corrected wavelength value is : 646.6

You might be interested in

What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not

mr_godi [17]

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

$F=G*\frac{m1*m2}{r^{2} }$

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

$F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }$

$F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }$

$F'=4*G*\frac{m1*m2}{r^{2} }$

F'=4*F

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

7 0

3 years ago

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2

lbvjy [14]

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

 = 3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

 Distance = (1/2) · (acceleration) · (time)²

 = (1/2) · (8 m/s²) · (30 sec)²

 = (4 m/s²) · (900 sec²)

 = 3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s². That's about 0.82 G !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.

5 0

3 years ago

A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6

Rina8888 [55]

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force $\frac{mv^{2} }{r}$ . Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

7 0

3 years ago

Please help calculate them A-f please very urganr

valentina_108 [34]

Three 40w lamp for 6 hours

8 0

4 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is 1215 N

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

1215 N is the force required to move the quarterback with linebacker.

5 0

3 years ago