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Lynna [10]
2 years ago
11

What net force is required to accelerate a car at a rate of 5 m/s ^ 2 if the car has a mass of 3, 000 kg ?

Physics
1 answer:
Olenka [21]2 years ago
3 0

Answer:

The answer is 15000kgms^-2

Explanation:

Actually here we are using the formula F=ma

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When water in clouds undergoes condensation, it can lead to?
olga_2 [115]
It will lead to rain. I know this because as the clouds move over water sources like oceans, lakes, and rivers, the water evaporates and rises. The water then liquefies into little water droplets. As the cloud moves over more water the droplets get scrunched up and get bigger over time and soon they get heavy and gravity pulls the droplets down to earth as rain. The End.    
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Label each formula in the chemical quation below as either a reactant of a product.
Karolina [17]
Fe and S are both reactants: they react with each other to give a different compound.

FeS is the product of the reaction: it was formed or produced as result of the reaction of Fe and S.

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Fe: reactant
S: reactant
FeS: product


3 0
3 years ago
A wave has an amplitude of 0.0800 m and is moving .33 m/s . One oscillator in the wave takes 0.115 s to go from the lowest point
garik1379 [7]

Answer:

0.230 s

Explanation:

The period is the length of time from one peak to the next.  If it takes the oscillator 0.115 s to go from the lowest point to the highest point, then it takes another 0.115 s to return to the lowest point.  So the period is 0.115 + 0.115 = 0.230 seconds.

6 0
3 years ago
2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas
Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

ds=\dfrac{dQ}{dt}

At constant pressure,

dQ=C_{p}dt

\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}

\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}

Put the value into the formula

14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})

\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}

0.693=ln\dfrac{T_{2}}{T_{1}}

ln2=ln\dfrac{T_{2}}{T_{1}}

T_{2}=2T_{1}...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

\Delta Q=C_{p}\Delta T

Put the value into the formula

6236=2.5\times8.3144(T_{2}-T_{1})

6236=20.786(T_{2}-T_{1})

T_{2}-T_{1}=\dfrac{6236}{20.786}

T_{2}-T_{1}=300

Put the value of T₂

2T_{1}-T_{1}=300

T_{1}=300\ K

Put the value of T₁ in equation (I)

T_{2}=2\times300

T_{2}=600\ K

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0
3 years ago
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