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3 years ago

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 v is the velocity of the wagons locked together, then

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

==> v ≈ 6.2 m/s

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

E / (2t ) = 1/2 × E/t = 1/2 P

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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

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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

Read 2 more answers

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

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

Travka [436]

1. Stop transmitting start and stop sending photons. This is typical morse code like transmissions of information. By varying the pattern of starts stops you vary the information sent. Also used in digital communications over radios, modems, and pulse dialing over radio/phones.

2. Change the frequency. You use a pattern of shorter and longer wavelengths of radio waves higher and lower energies of individual photons to transmit information. This is frequency modulation

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3 years ago