What information about an element is most crucial for locating that element in the periodic table?

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

2 years ago

Review:

Tresset [83]

Answer:

1. 1. A quantity is completely described by magnitude alone. A quantity Is completely described by a magnitude with a direction.

[a]. scalar, vector

b. vector, scalar

2.2. Speed is a velocity is a quantity and quantity.

a. scalar, vector

[b]. vector, scalar

3 0

1 year ago

A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of

Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0

2 years ago

Which of the following is an example of distance equaling displacement?

AleksandrR [38]

Answer:

D

Explanation:

The student had displaced their in the class when she left. The phone is what's displaced and student leaving equals distance.

3 0

2 years ago

Ok I am done waiting for an answer for 5 hours can somebody PLEASE PLEASE PLEASE help me TIMED ASSESMENT HERE!!!! PLEASE

JulsSmile [24]

Answer:

Explanation:

I'm going to assume that all the givens are in m/s. If I am correct, The orange line is speeding up. It is going from 4 m/s to 7 m/s. Its slant is from lower left to upper right. It is increasing in speed.

The green line is slowing down. It is going from 4 to 0 m/s. It's slant is from upper left to lower right.

The blue line is horizontal. It is neither slowing down or speeding up.

3 0

1 year ago

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