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3 years ago

A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the

Physics

1 answer:

olchik [2.2K]3 years ago

4 0

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

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Answer:

4 Ohms

Explanation

(This is seriously not as hard as it looks :)

You only need two types of calculations:

replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two: $R = R_1+R_2$
replace two resistances that are connected in parallel. In that case: $\frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_2}\\\mbox{or}\\R= \frac{R_1\cdot R_2}{R_1+R_2}$

I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the equivalent resistance of the original circuit.

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3 years ago

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jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

distance between the two black holes, r = 10 AU = 1.5 x 10¹² m
gravitational force between the two black holes, F = 6.9 x 10²⁵ N.
combined mass of the two black holes = 5.20 x 10³⁰ kg

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

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5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

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solve the quadratic equation using formula method;

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The second mass is calculated as follows;

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m₂ = 5.2 x 10³⁰ kg - 4.7 x 10³⁰ kg

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Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

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