2 years ago

Can someone please help me with this problem. I’ve tried but I can’t seem to figure it out.

Physics

1 answer:

Zolol [24]2 years ago

7 0

The last one, Tension = 30N, Normal force = 20 N

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Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once

GaryK [48]

NOTE: The diagram is attached to this solution

Answer:

The acceleration of point A = 14.64 ft/s²

Explanation:

By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)

$a_{A} = \bar{a} + \frac{\alpha L}{2}$

Weight, W = mg

g = 32.2 ft/s²

m = W/g

$p = m \bar{a}\\p = w \bar{a} /g$

$\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66$

$\frac{pL}{2} = I \alpha$

but $I = \frac{wL^{2} }{12g}$

$\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L$

$a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}$

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3 years ago

Calculate the acceleration due to gravity on the moon. the moon's radius is 1.74×106m and its mass is 7.35×1022 kg.

Kaylis [27]

184.44

7,511.7 that s the answer

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3 years ago

What is true about the inertia of two cars , car A of mass 1,500 kilograms and car B of mass 2,000 kilograms?

Elena-2011 [213]

Car B mass 2,000 kilograms

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3 years ago

The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

Julli [10]

Answer:

$v=(6ti+6k)\ m/s$

Explanation:

Given that,

The position of a particle is given by :

$r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m$

Let us assume we need to find its velocity.

We know that,

$v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s$

So, the velocity of the particle is $(6ti+6k)\ m/s$ .

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3 years ago

Select the correct answer

devlian [24]

Explanation:

please send full question....

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3 years ago

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