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Kitty [74]
2 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]2 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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The seasons on earth are caused by its elliptical orbit around the sun. the seasons on earth are caused by its elliptical orbit
Korolek [52]
That statement is false.
That statement is false.
7 0
3 years ago
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the specific heat of gold is 0.031 calories degrees Celsius and the specific heat of silver is 0.057 calories degrees Celsius so
givi [52]

Answer:

Explanation:

Given:

Specific heat of gold  = 0.031cal/°C

Specific heat of silver  = 0.057cal/°C

To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.

It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.

Now,

The higher the specific heat capacity the more energy it is required to heat up the substance.

So, Gold with a specific heat capacity of 0.031cal/°C  will heat up faster.

8 0
2 years ago
What is the peacocks displacement from 2 to 3 seconds
mash [69]
1.5 second difference
7 0
2 years ago
A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step
marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

=60 * 3

=180 kgm/s^2

=180 N

6 0
3 years ago
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
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