prepara un escrito en el que resumas las ideas principales sobre la evolución de los modelos atómicos

in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,

kirza4 [7]

The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( $R^{2}$ + $(R/2)^{2}$ )α

T.R = 1/2M 5 $R^{2}$ /4 α

T = 5Ma/8

Mg - T = Ma

Mg - 5Ma/8 = Ma

Mg= 5Ma/8 + Ma = 13Ma / 8

acceleration = 8g/13 = 6 m/s^2

The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

Learn more about acceleration here:

brainly.com/question/2303856

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4 0

1 year ago

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

finlep [7]

In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".

5 0

3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$