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Lostsunrise [7]

3 years ago

14

Light always follows one and the same path traveling back and forth. If a flashlight has a light bulb placed exactly in the foca

l point of its convex lens, it will produce a beam of..
A. converging light rays
B. diverging light rays
C. parallel light rays

1 answer:

Reil [10]3 years ago

5 0

Answer:

converging

.................

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lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of

bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

$U=\dfrac{1}{2}\epsilon_o E^2$

E is electric field

It means, $U\propto E^2$

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

$\dfrac{U}{U'}=(\dfrac{E}{E'})^2$

We have,

E'=2E

So,

$\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U$

So, the intensity of laser 2 is 4 times of the intensity of laser 1.

6 0

3 years ago

identified a problem in his/her community related to Earth Science. complete research to identify solutions to the problem, deve

son4ous [18]

Answer:

One of the best ways to gain insights in a Design Thinking process is to carry out some form of prototyping. This method involves producing an early, inexpensive, and scaled down version of the product in order to reveal any problems with the current design. Prototyping offers designers the opportunity to bring their ideas to life, test the practicability of the current design, and to potentially investigate how a sample of users think and feel about a product.

Prototypes are often used in the final, testing phase in a Design Thinking process in order to determine how users behave with the prototype, to reveal new solutions to problems, or to find out whether or not the implemented solutions have been successful. The results generated from these tests are then used to redefine one or more of the problems established in the earlier phases of the project, and to build a more robust understanding of the problems users may face when interacting with the product in the intended environment.

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7 0

3 years ago

3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

$F_{net}=ma$

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction $F_f$ , opposing the motion of the box, so to the left

So we can write the net force as

$F_{net}=F-F_f$

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

$a=0$

Therefore

$F_{net}=0$

WHich means:

$F-F_f=0$

And therefore,

$F_f=F=25 N$

which means that the force of friction is also 25 N.

6 0

3 years ago

How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to

jek_recluse [69]

Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days

6 0

3 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

3 years ago