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Zolol [24]
4 years ago
8

A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration

of 2.00 m/s² until its engines stop at an altitude of 150 m.
a. What is the maximum height reached by the rocket?
b. When does the rocket reach maximum height?
c. How long is the rocket in the air?
Physics
1 answer:
Nonamiya [84]4 years ago
3 0

Answer:

a)5m b)5secs c) 10secs

Explanation:

a) maximum height H = u²/2g

Given u = 50m/s

g = 10m/s²

H = 10²/20

H= 5m

maximum height reached by the rocket is therefore 5m

.b)Time it takes the rocket to reach its maximum height is Tmax = U/g

Tmax = 50/10

Tmax = 5seconds

c) How long the rocket spent in the air is the time of flight T

T = 2U/g

T = 2×50/10

T = 100/10

T = 10seconds

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3 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
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<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

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Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

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Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

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Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

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