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1 year ago

13

A 200 N object is resting on a surface with a coefficient of static friction equal to 0.15. A person pushes horizontally with a

force of 20 N. Calculate the magnitude of the force of friction.

1 answer:

Reika [66]1 year ago

3 0

Answer:

20 N

Explanation:

the force of static friction maximum = Fn * .15 = 200 * .15 = 30 N

the force of friction will cancel out the 20 N force applied

Ff = 20 N

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Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

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4 years ago

1. In what type of ecosystem can monkeys population be found?

sp2606 [1]

I, thank you for posting your question here at Brainly.

Here are the answers to your questions:

1. Monkeys live in tropical rainforest ecosystems.
2. Animals should be left alone in nature to take part in the balance of the ecosystem, and not upset it. A sudden increase of monkey population, probably due to destruction of rainforests which lead to lesser area of living for them.
3. Monkeys take a huge part in seed dispersal among flesh fruit plants. Without them, it will greatly affect the conservation of the forests.

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3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

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3 years ago

Two cylindrical resistors are made from same material and have the same length. When connected across the same battery, one diss

Ilia_Sergeevich [38]

Answer:

$\frac{d_2}{d_1} = \sqrt2 = 1.41$

Explanation:

As resistor is connected to the battery of constant EMF then the power across the resistor is given as

$P = \frac{E^2}{R}$

now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance

Due to different resistance they both will have different power

Since power is inversely depends on the resistance

So if the power is twice that of the other then the resistance must be half

so we have

$R_1 = \rho \frac{L}{A_1}$

$R_2 = \rho\frac{L}{A_2}$

since one resistance is half that of other resistance

So the area of one must be twice that of other

so we have

$\frac{A_2}{A_1} = 2$

$\frac{\pi d_2^2}{\pi d_1^2} = 2$

$d_2 = 1.41 d_1$

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3 years ago

Warm Front: advancing warm air displays cold air along a warm front develops a gradual boundary slope as an illustrated in figur

pishuonlain [190]

Answer:

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Explanation:

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