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ivolga24 [154]
2 years ago
12

A car traveling north with a velocity of 33 m/sec slows down to a velocity of 12 m/sec north within 10 sec. What is the car’s de

celeration?
Physics
2 answers:
Alja [10]2 years ago
6 0
33 m/s north is initial velocity (Vi). 12 m/s north is final velocity (Vf). 10s is the time. The equation for acceleration is Vf-Vi/t. 12-33 is -21. -21 divided by 10 is -2.1. The answer is -2.1 m/s^2 North.
strojnjashka [21]2 years ago
4 0

a=v/t

v= 33-12 = 21

21÷10=2.1

a= 2.1

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"The burning of fossil fuels produces gases which are capable of trapping heat resulting into the current rise in the global tem
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a. True

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Solar radiation at frequencies of visible light passes through the atmosphere, heating the planet's surface, subsequently this energy is emitted in infrared thermal radiation. This radiation is absorbed by the gases produced by the combustion of fossil fuels. Therefore, the greater the amount of these gases in the atmosphere, the more heat will be trapped in the earth, raising its global temperature.

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3 years ago
Two loudspeakers emit 600 Hz Hz notes. One speaker sits on the ground. The other speaker is in the back of a pickup truck. You h
Firdavs [7]

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The truck's speed is 4.04 m/s.

Explanation:

Given that,

Emit frequency = 600 Hz

Beat = 7.00 beat/sec

We need to calculate the truck's speed

Using formula of speed

\text{frequency observed}=\text{frequency emitted}\times\dfrac{v}{v+v_{source}}

Where, v = speed of sound

Put the value into the formula

(600-7)=600\times(\dfrac{343}{343-v_{truck}})

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2 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
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Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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