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Luda [366]
2 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]2 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]2 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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Sandy is whirling a ball attached to a string in a horizontal circle over his head. If Sandy doubles the speed of the ball, what
jeka57 [31]

The tension in the string B) It quadruples.

Explanation:

The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:

T= m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the ball

v is the speed of the ball

r is the radius of the circle (the lenght of the string)

In this problem, we are told that the speed of the ball is doubled, so

v' = 2v

Substituting into the previous equation, we find the new tension in the string:

T' = m \frac{(2v)^2}{r}=4(m\frac{v^2}{r})=4T

Therefore, the tension in the string will quadruple.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0
3 years ago
How do people participate in tourism
Makovka662 [10]
Community participation in tourism helps to uphold the local culture, tradition and indigenous knowledge of the local people. It also helps in conservation of the environment and culture of the local community.
5 0
2 years ago
The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

Mass of earth M_E=5.972\times 10^{24}kg

So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

(A) So the ratio of mean density \frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735

(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

5 0
3 years ago
Read 2 more answers
what is a hypothesis reffered to as after being verified by a large number or independent experiments
Iteru [2.4K]

Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

7 0
3 years ago
The diagram shows a position-time graph What is the displacement of the object
algol13

The object is moving, so at different times, it has different displacement.  I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the  beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) .  That's  <em>-1m </em>.


5 0
2 years ago
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