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MakcuM [25]
3 years ago
11

My Notes The linear density rho in a rod 5 m long is 11/ x + 4 kg/m, where x is measured in meters from one end of the rod. Find

the average density rhoave of the rod.
Physics
2 answers:
Lostsunrise [7]3 years ago
5 0
<h2>Corrected Question:</h2>

The linear density ρ in a rod 5 m long is (11x+4) kg/m, where x is measured in meters from one end of the rod. Find the average density ρ(ave)

<h2>Answer:</h2>

157.5kg/m

<h2>Explanation:</h2>

The linear density, ρ, of a rod is given by;

ρ = \frac{m}{l}

Where;

m = mass of the rod

l = length of the rod

The average density ρ(ave) is then, the average of the linear density taken over the entire length of the rod. i.e

ρ(ave) = \int\limits^a_b {p} \, dx            -----------------(ii)

Where;

a and b are the limits of the length of the rod.

b = 0

a = 5m

p = (11x + 4) kg/m         [as given in the question]

Substitute these values into equation (ii) as follows;

ρ(ave) = \int\limits^5_0 {(11x + 4)} \, dx

ρ(ave) = [\frac{11x^2}{2} + 4x ]₀⁵

Substitute the integral limits into the equation;

ρ(ave) =  [ \frac{11(5)^2}{2} + 4(5) ] - [\frac{11(0)^2}{2} + 4(0)]

ρ(ave) =  [ \frac{11(5)^2}{2} + 4(5)]

ρ(ave) =  [ 137.5 + 20)]

ρ(ave) =  157.5

Therefore, the average density of the rod is 157.5kg/m

dedylja [7]3 years ago
3 0

Answer:

ρ(ave) = 29.1 kg/m

Explanation:

The average value of a function over an interval = definite integral computed with the boundaries of the interval divided by the extent of the integral

ρ(ave) = [∫⁵₀ (11x + 4) dx]/(5-0)

ρ(ave) = [(11x²/2) + 4x]⁵₀ ÷ (5)

ρ(ave) = [(11(5²)/2) + (4×2)]/5

ρ(ave) = (137.5 + 8)/5

ρ(ave) = 29.1 kg/m

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A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
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Answer:

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Explanation:

From the question,

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W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

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Substitute equation 2 and equation 3 into equation 1

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Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

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4 0
3 years ago
In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0
3 years ago
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