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3 years ago

A rock climber weighs 550N and is carrying a backpack that weighs 50N. How much total work has been done after she climbs 12M

Physics

1 answer:

kherson [118]3 years ago

4 0

Answer:

7200 J

Explanation:

Work = force × distance

W = (550 N + 50 N) × (12 m)

W = 7200 J

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Find the total volume of the material used to making the cylinder

gladu [14]

Answer:

Hey mate

Explanation:

Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.

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2 years ago

A wild pig attacks a hunter with a velocity of 5 m/s. At the instant when the pig is at a distance of 100 m the hunter shoots an

castortr0y [4]

Answer:

31.55 m/s

Explanation:

Let the initial velocity of the arrow is u metre per second.

Angle of projection, θ = 40 degree

range = 100 m

Use the formula for the range.

$R = \frac{u^{2}Sin2\theta }{g}$

100 = u^2 Sin(2 x 40) / 9.8

100 x 9.8 = u^2 Sin 80

u^2 = 995.11

u = 31.55 m/s

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3 years ago

A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w

aliya0001 [1]

Answer:

f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

from work done = force x distance

work done = change in potential energy of the pile driver

equating the two equations above we have

force x distance = m x g x (s - d)

f x 0.12 = 2100 x 9.8 x (5- (-0.12))

d = - 0.12 because the steel beam went down at we are taking its

initial position to be an origin point which is 0

f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

f = 878,080 N

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3 years ago

Under which conditions are particles in a medium said to be in phase with one another?

rewona [7]

Answer:

Corvette Corvette

Explanation:

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The answer is A btw :)

7 0

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.

7 0

3 years ago