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1 year ago

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A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea

ches the lowest point on its path, what is the maximum height of the pendulum?

2 answers:

Scrat [10]1 year ago

4 0

Answer : Maximum height of the pendulum is 0.43 m

Explanation :

Given that,

Mass of the pendulum, m = 2 kg

Speed of the pendulum, v = 2.9 m/s

When it reaches the lowest point on its path, the potential energy at the highest point is equal to the kinetic energy at the lowest point.

i.e.

$KE=PE$

$\dfrac{1}{2}mv^2=mgh$

h is the height of the pendulum.

$h=\dfrac{v^2}{2g}$

$h=\dfrac{(2.9\ m/s)^2}{2\times 9.8\ m/s^2}$

$h=0.43\ m$

Hence, this is the required solution.

Vanyuwa [196]1 year ago

3 0

You first us 1/2(mv^2) to solve for the potential energy and then put that in to PE=m*g*h and solve for hight

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Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

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1 year ago

A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

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2 years ago

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2

Ostrovityanka [42]

For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then
$d ^ 2 = 16 ^ 2 + 12 ^ 2
$
From here, we clear the value of d.
We have then:
$d = \sqrt{16 ^ 2 + 12 ^ 2} 
$
Rewriting:
$d = \sqrt{256 + 144}$
$d = \sqrt{400}$
$d = 20 miles
$
Answer:
The magnitude of the car's displacement is:
d = 20 miles

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1 year ago

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Dafna11 [192]

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>

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1 year ago

A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.

Anna007 [38]

Answer:

$\mu_k = \frac{2(vt - L)}{(g + a) t^2}$

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

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$a = -\mu_k(g + a)$

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1 year ago