PHYSICS COLLEGE

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answers

Answer 1

Answer:

Answer:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \nVector (OB) = 0i + 3j + 2k \nVector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_(AB) ,u_(AC), u_(AD), u_(AE)$ as follows:

$u_(AB) = (vector(AB))/(magnitude(AB)) \n= \frac{OB - OA}{magnitude({vector(OB - OA))} }\n=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \n\n=-0.857 i +0.429j+0.286k\n\nu_(AC) = (vector(AC))/(magnitude(AC)) \n= \frac{OC - OA}{magnitude({vector(OC - OA))} }\n=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \n\n=-0.857 i -0.286j+0.429k\n\nu_(AD) = +1i\nu_(AC) = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\nF_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\nF_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\nW = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\nF_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\n(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i) + (-160k) = 0$

Comparing i , j and k components as follows:

$-0.857F_(AB) -0.857F_(AC) +1F_(AD) = 0\n+ 0.429F_(AB) - 0.286F_(AC) = 0\n+0.286F_(AB) +0.429F_(AC) = 160$

Solving Above equation simultaneously we get:

$F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004$

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You have 5 cats and they each have a mass of 4kg per cat. What is the mass of all of them together?

Answers

Answer:

It would be 20kg

Explanation:

This would be just 5x4 as there are 5 cats and each are 4kg. You can also add 4, 5 times as well.

I hope Im correct

Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 11.0 cycles/s (Hz). What is the maximum speed ( v ) of the object?A. 165 m/s
B. 1.65 m/s
C. 10.4 m/s
D. 1040 m/s

Answers

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.

Answers

Answer:

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600 seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

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