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DanielleElmas [232]

3 years ago

6

14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 65.0°-angle with the horizontal. (a)

Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.10 m along the ladder from the bottom. horizontal

1 answer:

QveST [7]3 years ago

4 0

Answer:

The horizontal reaction force is 230.3 N.

Explanation:

Given that,

Length l= 14.0 m

Weight of ladder F = 490 N

Angle = 65°

Weight of firefighter F'= 850 N

Height l'= 4.10 m

Suppose horizontal force magnitude N direction vertical force magnitude N direction

We need to calculate the horizontal reaction force

Horizontal reaction force = normal reaction from wall

Vertical reaction force = weight of ladder +weight of man

$F=490+850$

$F=1340\ N$

We need to calculate the moment about bottom is zero

$N\times l\sin\theta=F\times\dfrac{l}{2}\cos\theta+F'\times l'\cos\theta$

Put the value in the equation

$N\times14\sin(65)=490\times7\cos(65)+850\times4.10\cos(65)$

$N=\dfrac{490\times7\cot(65)+850\times4.10\cot(65)}{14}$

$N=230.3\ N$

Hence, The horizontal reaction force is 230.3 N.

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Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

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using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

<u>D = - 0.57</u>

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

<u>C = 2.43</u>

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3 years ago

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Invasive Species

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3 years ago

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

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2 years ago

How many Mars-sized planets can fit in Jupiter

Maru [420]

Answer:

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Explanation:

7 0

3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago