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Luda [366]
3 years ago
7

Which would most likely cause specular reflection? O a pathway with rough rocks a shiny, smooth leaf a small patch of soil a rou

gh log​
Physics
2 answers:
andrew11 [14]3 years ago
8 0

Answer:

a shiny smooth leaf

Explanation:

A shiny smooth leaf will cause specular reflection. Other choices will cause diffused reflection from the surface.

A specular reflection is similar to how a mirror or smooth surface reflects. The incident light is given off as a single ordered reflection from the surface of a body.

For this to occur, the surface incident must be smooth and without rough patterns on it.

A path way with rough rocks, small patch of soil and rough logs will give off diffused reflection

qwelly [4]3 years ago
7 0

Answer:

B

Explanation:

that is the answer for shiny smooth leaf on my edg but he is right

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In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?
miskamm [114]

<u>Answer:</u>

<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.

7 0
3 years ago
19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged
Nutka1998 [239]
<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F 

So: 

F' = 8F 

This implies that the force that you would feel pulling you down (your weight) would increase by 800%! 

You would be 8 times heavier on this "new" earth!</span>
4 0
3 years ago
2. Draw a ray diagram that shows the image
Lunna [17]

you have to draw diagram

6 0
3 years ago
A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

          a = (0 i ^ - g j⁾

          a = (0i ^ - 9.8 j ^) m / s²

5 0
3 years ago
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