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Darya [45]
3 years ago
14

Viteza unui mobil care se deplasează cu accelerație constantă crește de la 3,2 m/s la 5,2 m/s în timp de 8 s. Accelerația mobilu

lui este :
Engineering
1 answer:
Verizon [17]3 years ago
3 0

Answer:

a=0.25\ m/s^2

Explanation:

Initial speed of the mobile = 3.2 m/s

Final speed of the mobile = 5.2 m/s

Time, t = 8 s

We need to find the acceleration of the mobile. It can be given by the change in velocity divided by time. So,

a=\dfrac{v-u}{t}\\\\a=\dfrac{(5.2-3.2)\ m/s}{8\ s}\\\\=0.25\ m/s^2

So, the acceleration of the mobile is 0.25\ m/s^2.

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An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
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3 0
3 years ago
In response to the market revolution:the legal system worked with local governments to find better ways to regulate entrepreneur
Nataliya [291]

Answer:

Local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.

Explanation:

The Market Revolution is the name given to change in the economy that occurred in the 19th century. This drastic change led to various important changes in the United States and across the world. During this period, capitalism became more entrenched and society became, for the first time, predominantly capitalist. This gave businessmen great power, as they played an increasingly important role when it came to economic growth. The power that they had influenced society deeply, including legislation. Judges often protected businessmen from paying property damages that were associated with their business enterprises. Moreover, workers had few rights and protections, and judges prevented them from unionizing.

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3 years ago
What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin
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Answer:

m_{LP}=0.45\,kg

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]

Q_{water} = 3599.435\,kJ

The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

Q_{LP} = 22496.469\,kJ

A kilogram of LP gas has a minimum combustion power of 50028\,kJ. Then, the required mass is:  

m_{LP} = \frac{22496.469\,kJ}{50028\,\frac{kJ}{kg} }

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6 0
3 years ago
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