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Pavel [41]
3 years ago
10

A bucket holding a cartoon character is dangling from a rope above a well from the windlass which raises the bucket from the wel

l. The bucket and character have a combined mass of 10.0 kg. The distracted friend lets go of the handle, and the bucket falls, unwinding the windlass. The acceleration of the character and bucket is 4.9 m/s2. The windlass has a radius of 0.20 meters. What is the moment of inertia of the windlass
Physics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

   I = 0.4 kg m²

Explanation:

Let's solve this exercise in part, let's first work with the cube.

      W -T = m a

      T = W - m a

      T = m (g-a)

      T = 10.0 (9.8 - 4.9)

      T = 49 N

Now let's work on the pinwheel, we write Newton's rotational law

     τ = I α

     T r = I α

Linear and angular variables are related.

     a = α r

     α = a / r

We replace

    T r = I a / r

    I = T r² / a

we calculate

   I = 49 0.20² / 4.9

   I = 0.4 kg m²

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3 years ago
In multicellular organisms, which choice ranks the level of organization from simplest to increasingly more complex?
Andreyy89

Answer:

<em>The answer is C.</em>

Explanation:

Cells are the smallest organisms and makes up all living thing or being. The tissues are made of cells. The organs are made of tissues. The organ systems are made of organs. Therefore the order is C. cells, tissues, organs, and organ systems.

3 0
3 years ago
Practice physics chapter 3 grade 11 test
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6 0
4 years ago
Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3
Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N

F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}

The net force on q3 is the sum of these two forces:

F_{net} = F_{13} + F_{23} = -1.92\times 10^{-5} + (-19.5\times10^{-6}) = 2.8\times 10^{-5} N

Since both forces are directed towards left, their sign should be negative.

5 0
3 years ago
In moving out of a dormitory at the end of the semester, a student does 2.17 x 104 J of work. In the process, his internal energ
galina1969 [7]

Answer:

a. W = 2.17 x 10 ⁴ J

b. ΔU = - 5.03 x 10⁴ J

c. Q = 2.86 x 10 ⁻³ J

Explanation:

a.

The student work does 2.17 x 10⁴ J

So W = 2.17 x 10 ⁴ J  

b.

The internal energy decreases 5.03 x 10⁴ J

ΔU = U₁ - U₂

ΔU = 0 - 5.03 x 10⁴ J = - 5.03 x 10⁴ J

c.

The formula heat can use:

Q = W + ΔU

Q = 2.17 x 10⁴ J + ( - 5.03 x 10⁴ J )

Q = 2.86 x 10 ⁻³ J

4 0
3 years ago
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