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Mrac [35]
4 years ago
14

How fast is an airplane traveling that accelerates at a rate of 20 mph per second for 30 seconds?

Chemistry
2 answers:
Basile [38]4 years ago
8 0

Explanation:

The given data is as follows.

                   acceleration (a) = 20 mph

                   time (t) = 30 sec

                  initial velocity (u) = 0

As 1 mph = 0.447 m/s. So, convert 20 mph into m/s as follows.

               20 mph/s \times \frac{0.447 m/s}{1 mph}

                   = 8.94 m/s^{2}

Relation between velocity, acceleration and time is as follows.

                            v = u + at

Putting the given values into the above formula as follows.

                          v = u + at

                          v = 0 + 8.94 m/s^{2} \times 30 sec

                             = 268.2 m/s

Therefore, we can conclude that at a speed of 268.2 m/s the airplane is travelling in the given situation.

Nostrana [21]4 years ago
6 0
Should be 600! hope this helps.
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hurry please! avogadro's law relates the volume of a gas to the number of moles of gas when temperature and pressure are constan
DIA [1.3K]

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:

V1/n1 = V2/n2

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V x n2 = 2 x 2V

Divide both side by V

n2 = (2 x 2V)/V

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5 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

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ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

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L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

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Answer:

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