All categories

1 year ago

Ce unitate de masura are indicele de REFRACTIE?

1 answer:

3 0

Answer:se obesrva ca indicele de refractie NU are unitate de masura, este adimensional. Se exprima print-un raport, de exemplu 4/3, 1,5 etc.

Explanation:

You might be interested in

In a thundercloud there may be an electric charge of 41 C near the top of the cloud and −41 C near the bottom of the cloud. If t

Andre45 [30]

Answer:

F = 236063.6N

Explanation:

Please see attachment below.

4 0

2 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t + 1/2 a t²

s = 1/2 a t² ----------- (2)

compare equation (1) and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0

2 years ago

When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

3 0

2 years ago

A race car starting from rest accelerates uniformly at a rare of 4.90 meters per second^2. What is the cars speed after it has t

Vesnalui [34]

From the law of Galileo Galilei :v²=v₀²+2ad we take the speed

v²=0+2*4.90*200=1960=>v=√1960=44.27 m/s

4 0

2 years ago

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation