Answer:
B. The water
Explanation:
Water is abiotic factor because it is non living
The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.
<h3>What is kinetic energy?</h3>
- A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
- Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
- The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
- Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.
The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.
To learn more about kinetic energy, refer to:
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What Is A Molecule? B) NaCi
Answer:
The correct answers are
(a) It decreases to 1/3 L
(ii) is (c) It is constant
Explanation:
to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem
The given variables are as follows
Initial volume V1 = 1L
V2 = Unknown
Initial Temperature T1 = 300K
let us assume that the balloon is perfectly elastic
At 300K the balloon is filled and it stretches to maintain 1 atmosphere
at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,
For (i) since we know that the pressure of the balloon is constant
by Charles Law V1/T1 =V2/T2
or V2 = (V1/T1)×T2 =
×
=
× L = L/3 hence the correct answer to (i) is 1/3L
Answer:
The COP of the system is = 4.6
Explanation:
Given data
Higher pressure = 1.8 M pa
Lower pressure = 0.12 M pa
Now we have to find out high & ow temperatures at these pressure limits.
Higher temperature corresponding to pressure 1.8 M pa
°c = 335.9 K
Lower temperature corresponding to pressure 0.2 M pa
°c = 262.9 K
COP of the system is given by


COP = 4.6
Therefore the COP of the system is = 4.6