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SVETLANKA909090 [29]
3 years ago
11

Clouds form from _____ temperature change, which occurs when an expanding gas cools.

Physics
1 answer:
IrinaVladis [17]3 years ago
8 0
The answer is Adiabatic
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NO LINKS! ASAP PLEASE *FOR TEST TOMORROW*
denis23 [38]

Answer:

Friction: When two surfaces slide against each other / make contact with each other

Force: A push or pull exerted on an object

8 0
3 years ago
while performing a sonogram, to minimize risk to the fetus, which imaging mode trade off would be most appiclable.
Alexus [3.1K]

Answer: Use M-mode instead of pulse-wave Doppler to measure the fetal heart rate.

Explanation: M-mode is a famous ultrasound process during the early days of Ultrasound technologies, M-mode has a very high sampling rate. One of the major disadvantages is that it may be difficult to align the M-mode to be the perpendicular to the SEPTUM. M-mode has only one (1) dimensional view of an object,to minimize the risk to the Fetus, M-mode is the most applicable tradeoff. Specifically, M-mode is mainly used because of its Axial and temporal resolution of the structure of objects.

7 0
4 years ago
Mali loves to make herself dizzy. She spins in place 7 times before falling down right where she was standing. Find her distance
Mila [183]
The answer is 0 bc she didint move
Explanation:
8 0
3 years ago
Read 2 more answers
If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.
Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so \frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{P_1}{P_2}=\frac{T_1}{T_2}

As T_1 is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0
3 years ago
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
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