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MatroZZZ [7]
4 years ago
8

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]4 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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6 0
3 years ago
A 10.3 kg weather rocket generates a thrust of 240 N. The rocket, pointing upward, is clamped to the top of a vertical spring. T
jenyasd209 [6]

Answer:

(a) x = 0.25 m

(b) v = 1.46 m/s

(c) v = 2.4 m/s  

Explanation:

mass (m) = 10.3 kg

force from thrust (F) = 240 N

spring constant (k) = 400 N/m

stretch distance from thrust (y) = 30 cm = 0.3 m

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

(A) from mg = kx

compression (x) = mg/ k

x = \frac{10.3 x 9.8}{400}

x = 0.25 m

   

(B) from the conservation of forces

  (Fy) + (0.5kx^{2}) =  (0.5ky^{2}) + mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}) -  (0.5ky^{2}) - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}) -  (0.5 x 400 x 0.3^{2}) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 1.46 m/s

                 

(C) if the rocket weren't attached to the spring, the conservation of energy equation becomes

 (Fy) + (0.5kx^{2}) = mgh + (0.5mv^{2})

v = \sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2})  - mgh }{0.5m}[/tex]}[/tex]

v =  \sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2})  - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]

v = 2.4 m/s                          

7 0
3 years ago
Wolf spiders carry their young on their backs. How does this help in their survival?
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It intimidates other spider because it makes them look bigger than they really are.
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When it gets that hot the evergreens can explode?
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If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk
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3 years ago
Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,
NikAS [45]
Kepler's 3rd law is given as
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where
P = period, days
A = semimajor axis, AU
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Given:
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Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer:  1.3439 x 10⁵ (days²/AU³)

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3 years ago
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