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4 years ago

1. If a car is traveling 90 mi/hr, how many feet will the car travel in 1 sec? In 5 sec?

Chemistry

1 answer:

Fudgin [204]4 years ago

3 0

Answer:

The car travel 660 feet

Explanation:

First convert the Speed into Feet/sec

Speed of the car = 90 mi/hr

1 mi = 5280 ft

In 90 miles = 90 x 5280 = 475,200 feet

1 hr = 60 min

1 min = 60 sec

So ,

1 hr = 60 x 60 sec = 3600 sec

The speed is defined as the distance traveled by the object in unit time. The formula of speed is :

$Speed =\frac{distance}{time}$

$Speed =\frac{90mi}{1hr}$ ... given

$Speed =\frac{475200 ft}{3600sec}$

Speed = 132 ft/sec

Now, time = 5 sec

$Speed =\frac{distance}{time}$

$distance =speed\times time$

$distance = 132ft/sec\times 5sec$

Distance = 660 feet

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A racing car travels 340 miles in 120 minutes. what speed is traveling at?

Papessa [141]

Answer:

It is traveling at 170 miles per hour.

Explanation:

divide 340 by 2 because right now it's at 340 miles per two hours.

You get 170 miles per hour.

4 0

3 years ago

What transition energy corresponds to an absorption line at 527 nm?

Ede4ka [16]

Answer:

E = 3.77×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of absorption line = 527 nm (527×10⁻⁹m)

Energy of absorption line = ?

Solution:

Formula:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = speed of wave = 3×10⁸ m/s

by putting values,

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m

E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m

E = 0.0377×10⁻¹⁷ J

E = 3.77×10⁻¹⁹ J

4 0

3 years ago

Read 2 more answers

Need chem help asap

gavmur [86]

Oh I’m so4+ds2=563

Which is the answer

5 0

3 years ago

What is the freezing point depression of a solution that contains 0.705?

Sveta_85 [38]

Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(freezing point) = (Kf)m
ΔT(freezing point) = 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C

Hope this answers the question. Have a nice day.

6 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

4 years ago