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eimsori [14]
3 years ago
10

A car accelerates at a rate of 5ft/s/s for a time of 9 seconds. How far does the car go?

Physics
2 answers:
quester [9]3 years ago
7 0

Answer: The car will move in a speed of 45 meter per second

Explanation:

vlada-n [284]3 years ago
4 0

The car will move in a speed of 45 meter per second

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These questions !plz !! i need help!!!
Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then

(140 kg) (15 m/s) = (140 kg + 200 kg) <em>v</em>

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(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:

<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>

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3 years ago
A 60-kg man uses an electric chair to carry him up the stairs. if the chair has a mass of 200 kg, what power must a motor genera
mrs_skeptik [129]
1200w that is the answer I don't really know
4 0
3 years ago
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The elements that do not ordinarily form compounds are
ololo11 [35]

The noble gasses, they are the elements on group 18 of the periodic table and do not usually form compounds. The noble gases are as follows: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

Hope this helped :)

 

8 0
3 years ago
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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.

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3 years ago
Two ways that radio waves are used for transmitting information. Help me please.
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