I NEED THE ANSWER QUICK PLEASEE

shtirl [24]

Since each serves a different purpose, theories cannot become laws. Explaining how or why a natural phenomenon occurs is what the set of ideas called theories do. On the other hand, mathematical relationships that describes what happens are what is done by laws.

Let me show an example that illustrates the points. Describing what happens in the natural world are done by the mathematical formulas called the Gas Laws. In this example, it would show that by using the Gas Laws, I will be able to predict with great accuracy the pressure if I double the temperature of a sealed gas. This idea is a law since the relationship is mathematical and it tells us what will happen.

<span>On the other hand, in order to explain why gases behave like the way they do, we must use the kinetic molecular theory.</span>

7 0

3 years ago

a professional athlete who takes dance or gymnastics to improve their skills in their professional sport.

scoundrel [369]

Yes, many professional athletes will take dance classes to improve their skills. Such as football players to improve their footwork and use other muscles they wouldn't use while doing just push ups.

4 0

4 years ago

An electric lightbulb in a lamp fixture is illuminating a small room. The system consists of the lightbulb, the lamp fixture, th

mixas84 [53]

The role of thermal energy is to balance the electrical energy lost from the light bulb.

<h3>Energy conservation</h3>

The law of energy conservation states that energy can neither be created nor destroyed but can be transformed from one form to another.

<h3>Thermal energy</h3>

This is a form of energy that occurs due to rise in temperature.

The role of thermal energy is to balance the electrical energy lost from the light bulb.

Learn more about conservation of energy here: brainly.com/question/166559

6 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago

Ball a of mass 5.0 kilograms moving at 20 meters per second collides with ball b of unknown mass moving at 10 meters per second

RUDIKE [14]

M = mass of the ball A = 5.0 kg

m = mass of the ball B = ?

V = initial velocity of the ball A before collision = 20 m/s

v = initial velocity of the ball B before collision = 10 m/s

V' = final velocity of the ball A after collision = 10 m/s

v' = final velocity of the ball B after collision = 15 m/s

using conservation of momentum

M V + m v = M V' + m v'

(5.0) (20) + m (10) = (5.0) (10) + m (15)

100 + 10 m= 50 + 15 m

5 m = 50

m= 10 m/s

4 0

3 years ago