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vredina [299]
2 years ago
10

A concrete highway curve of radius 60.0 m is banked at a 19.0 ∘ angle. what is the maximum speed with which a 1400 kg rubber-tir

ed car can take this curve without sliding? (take the static coefficient of friction of rubber on concrete to be 1.0.)
Physics
1 answer:
allochka39001 [22]2 years ago
7 0

Answer:24.26m/s

Explanation:coefficient of static friction is = v^2/rg

1.0=v^2/60*9.81

1.0=v^2=588.6

V^2=588.6

V=24.26m/s

Guest
1 year ago
looooooooooooooooooooooooooser.
Guest
1 year ago
sorry. I did not mean that. Your answer of 24.26m/s is wrong.
Guest
1 year ago
My apologies.
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Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment
riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = \frac{360+240}{450}

tan θ  = \frac{600}{450}

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

M__B}=(Fcos \theta *240)-(Fsin \theta *450)

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)

M__B}= 194400.463-485999.348

M__B}=-291598.885 N-mm\\

M__B}=-291.598 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B (M__B}),

we have:

M__B}=-Fcos \theta *360 - Fsin \theta * 0

M__B}=-Fcos \theta *360 - 0

M__B}=-Fcos \theta *360

where Force(F) =  1350 N and θ  = 53.13° ; we have:

M__B}= -1350*cos53.13^0*360

M__B}= -291600 N-mm

M__B}= -291.6 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

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4. In a closed system consisting of a cannon and a cannonball, the kinetic energy of a cannon is 72,000 J. If the cannonball is
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Answer:

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Explanation:

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Answer:

c>d>f=a>b>e

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When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

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Answer:

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Then, x_all = 150cm + 140cm = 290cm

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