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Dahasolnce [82]
3 years ago
8

Which is an application of the principle of uniformitarianism?

Physics
1 answer:
Allisa [31]3 years ago
3 0
"Studying a modern river valley to learn about how ancient rock layers formed" is the one among the following choices given in the question that is an <span>application of the principle of uniformitarianism. The correct option among all the options that are given in the question is the third option or option "C". </span>
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Plz help me thank you
jarptica [38.1K]
Area is how much matter somethin has hypothetically speaking if u take one cycinder and put it in a squad purism with lengths of .52 the area will be equal to pi
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2 years ago
How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the
VLD [36.1K]

Answer:

The required work done is 2555.448~J

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '\mu_{k}' be the coefficient of friction, then

f = \mu_{k} \times N = \mu_{k} \times Mg

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N

So the work done (W) in moving the crate by a distance s = 10.6 m is

W = F \times s = 241.08~N \times 10.6~m = 2555.448 J

5 0
3 years ago
Which of the following waves have the longest wavelengths?
Paladinen [302]
Radio waves have the longest wavelengths
7 0
3 years ago
Read 2 more answers
At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
 The value of the acceleration is:
 a = -g = -9.8 m / s ^ 2
 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
 For t = 5.8 we have:
 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
 (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
 h '(t) = -9.8 * t + 28.42
 -9.8 * t + 28.42 = 0
 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
8 0
3 years ago
Read 2 more answers
Write relationship between hertz and megahertz​
dusya [7]

Explanation:

1 mega Hertz = 1000000 hertz

8 0
2 years ago
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