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Kitty [74]
3 years ago
15

The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a ___________ and was often cov

ered in elaborate carvings
Physics
1 answer:
Ket [755]3 years ago
5 0

Answer: Tympanum

Explanation: A Tympanum could be described as the semicircular space above the doorway or window or entrance of a building. The space is usually designed using elaborate sculptural designs, Rocky costumes and ornaments. The tympanum could also be triangular-shaped and attributed to classical Greece and Roman architecture. It is usually located between the portals archivolt and the lintel. The tympanum is commonly observed in the churches and temples of ancient Rome and Greece.

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Waves of energy in the electromagnetic spectrum travel through space and into the earths atmosphere. Which waves pass more crest
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The answer is D.......
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What conditions are typcial of the soil the deeper we dig into it?
Tems11 [23]
When we first dig into soil, it's nice and soft.( Depending on what soil it is; usually topsoil) 

But when we go deeper into it, there are rocks and the soil is harder. There is bedrock at the bottom, subsoil in the middle, and topsoil at the top, which is where we walk.
8 0
3 years ago
A scientist conducting a field investigation records measurements of very low pressure and high relative humidity at the top of
Ivan

Answer:

Please find the answer in the explanation

Explanation:

Given that a scientist conducting a field investigation records measurements of very low pressure and high relative humidity at the top of a mountain.

Since a weather map indicates that a warm front is approaching the mountain, according to the conventional current, the warm front is approaching because the weather must have been in higher relative humidity in cool air.

The warm front is approaching to replace it so that the cool air can conventionally replace the warmth air too.

The condition the scientists will most likely observe at the top of the mountain will be high relative humidity.

5 0
3 years ago
A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c
Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; m_{totalv

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg +  0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

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2 years ago
Which term refers to the upper part of a glacier that receives the most snowfall?
liq [111]
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