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2 years ago

Research has shown that the electrons orbit the nucleus in circular motions as shown on the classic Bohr model.

Physics

2 answers:

Stels [109]2 years ago

6 0

Research has shown that the electrons orbit the nucleus in circular motions as shown on the classic Bohr model. FALSE

astraxan [27]2 years ago

4 0

False. The Bohr Model is fundamentally incorrect. Electrons do not travel along specific paths.

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Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra

Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

$\alpha = \frac{\tau}{I}$

The moments of inertia of each element are described below:

Hoop

$I = M\cdot R^{2}$

Solid sphere

$I = \frac{2}{5}\cdot M \cdot R^{2}$

Flat disk

$I = \frac{1}{2}\cdot M \cdot R^{2}$

Hollow sphere

$I = \frac{2}{3}\cdot M \cdot R^{2}$

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0

3 years ago

Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in

m_a_m_a [10]

Gas because liquids and solids volumes don't change from switching containers.

3 0

2 years ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello

klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0

3 years ago

A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the

shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

$Force, F = ma$

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

$a = 0.47 * 9.8 m/s^2$

$a = 4.606m/s^2$

We know,

$v^2 - u^2 = 2as$

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

$u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s$

The car travels at a speed of 25m/s.

6 0

3 years ago