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2 years ago

Rotational dynamics physics question please help.

Physics

1 answer:

soldier1979 [14.2K]2 years ago

8 0

Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points

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The IMA of the pulley show is

Lera25 [3.4K]

The IMA of the pulley shown is 2.

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2 years ago

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Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of

Sonbull [250]

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

a = 4 m , b = 4 m , c = 5 m

a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

bx−ay=0

Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

bx+2ay=ab

Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

2bx+ay=ab

Solve all three equations simultaneously:

bx−ay=0 , bx = ay

ay + 2ay = ab , 3ay = ab , y = b/3

bx = b/3

x = a / 3

So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

[ 2a/3 , b/3 ]

[ 2.67 , 1 ] m

3 0

2 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$