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kipiarov [429]
3 years ago
14

What is the minimum time that a warm up should last

Physics
2 answers:
Vladimir [108]3 years ago
7 0
5 minutes is the minimum time
djyliett [7]3 years ago
7 0
At minimum, a warm up stretch or exercise should last about 5 minutes. this way your muscles are well prepared 
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Suppose that the Hubble Space Telescope discovers a series of planets with the following characteristics moving around a star th
nalin [4]

Answer:

Option (D)

Explanation:

Terrestrial planets refers to those four planets that are nearest to the sun and that lies within the asteroid belt. These planets are mainly composed of rocks or other metal objects that have a hard and resistant surface on it. They have a metal core that is molten (liquid) in nature, and atmosphere is relatively less dense, and also various geological features are present on it like the crater, volcanoes which can be observed with the help of satellites. The average densities of these planets is about four times the density of water. For example, the density of water is 1 g/cm³, whereas the density of earth is 5.5 g/cm³.

Thus, the correct answer is option (D).

3 0
3 years ago
In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w
VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

8 0
3 years ago
Sally and Juan’s science teacher has directed the class to build a paper airplane. The challenge is to build an airplane that fl
devlian [24]
Evaluating a solution
8 0
3 years ago
Read 2 more answers
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
Please help question in photo
bonufazy [111]

Answer:

A.) the car

Explanation:

hope this helped <3

8 0
3 years ago
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