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snow_lady [41]
3 years ago
13

Solve the attachment...​

Mathematics
2 answers:
sasho [114]3 years ago
6 0

Answer:

2 ( Option A )

Step-by-step explanation:

The given integral to us is ,

\longrightarrow \displaystyle \int_0^1 5x \sqrt{x}\ dx

Here 5 is a constant so it can come out . So that,

\longrightarrow \displaystyle I =  5 \int_0^1 x \sqrt{x}\ dx

Now we can write √x as ,

\longrightarrow I = \displaystyle 5 \int_0^1 x . x^{\frac{1}{2}} \ dx

Simplify ,

\longrightarrow I =  5 \displaystyle \int_0^1 x^{\frac{3}{2}}\ dx

By Power rule , the integral of x^3/2 wrt x is , 2/5x^5/2 . Therefore ,

\longrightarrow I = 5 \bigg( \dfrac{2}{5} x^{\frac{5}{2}} \bigg] ^1_0 \bigg)

On simplifying we will get ,

\longrightarrow \underline{\underline{ I = 2 }}

Snezhnost [94]3 years ago
4 0

Answer:

Step-by-step explanation:

=  \int\limits^1_0 {5x\sqrt{x} } \, dx

=  \int\limits^1_0 {5xx^{1/2} } \, dx

= \int\limits^1_0 {5x^{3/2} } \, dx

= 5 \int\limits^1_0 {x^{3/2} } \, dx

= 5*\frac{2}{5}*x^{5/2}  |\left[\begin{array}{ccc}1\\0\\\end{array}\right] \left

= 5*\frac{2}{5}*1^{5/2}

= 2

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Answer:

  4b: 8.6 m/s²; 1.3×10^5 N; 2.1×10^4 N decrease

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Step-by-step explanation:

<h3>4b</h3>

i) The acceleration due to gravity is inversely proportional to the square of the distance between the objects. The distance to the shuttle is ...

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Then the acceleration due to gravity at the height of the space shuttle is about ...

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<h3>5c</h3>

i) The gravitational potential energy is given by ...

  U = -GMm/r

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ii) At a height of 3×10^4 m, the denominator in the above expression changes from 6.4×10^6 to 6.43×10^6. This changes the gravitational potential energy by a factor of 6.4/6.43 to -1.6×10^12 J

(Note: we're carrying only 2 significant figures in the result in accordance with the rules for precision in such calculations. The change is noticeable at the level of the 4th significant figure, less than 1/2%.)

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