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s344n2d4d5 [400]
3 years ago
6

the law of reflection states that the angle of incidence is equal to the angle of reflection. true or false.

Physics
1 answer:
mina [271]3 years ago
7 0
Yes it is absolutely true!
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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

8 0
3 years ago
Why do scientists believe that dark matter exists even though it cannot be seen?
poizon [28]
Because although they cannot see it, they can see it's influence on objects that can be seen, and it's effects.
3 0
2 years ago
A train traveled 350km in 2.5h. what was the average speed of the train?
MaRussiya [10]
Speed equals distance divided by time, so 350 divided by 2.5 equals 140 kilometers per hour.
3 0
3 years ago
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Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with
Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}

g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}

6 0
3 years ago
Electric power plants are the main contributors of what? Oxygen pollution, Sulfur dioxide pollution, thermal pollution, mercury
zloy xaker [14]
The correct answer is sulfur dioxide pollution

Hope this helped :)
6 0
3 years ago
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