During a race there are lots of forces that are exerted on a race car. One of these forces is friction.

Write the limitation of Rutherford atomic model

EastWind [94]

Explanation:

Rutherford's model was unable to explain the stability of an atom. According to Rutherford'spostulate, electrons revolve at a very high speed around a nucleus of an atom in a fixed orbit. However, Maxwell explained accelerated charged particles release electromagnetic radiations.

5 0

3 years ago

Situation 6.1 A 13.5-kg box slides over a rough patch 1.75 m long on a horizontal floor. Just before entering the rough patch, t

GenaCL600 [577]

B) 14.0 N

The way to solve this problem is to determine the kinetic energy the box had before and after the rough patch of floor. The equation for kinetic energy is:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Substituting the known values, let's calculate the before and after energy.

Before:

E = 0.5 M V^2

E = 0.5 13.5kg (2.25 m/s)^2

E = 6.75 kg 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 13.5kg (1.2 m/s)^2

E = 6.75 kg 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

So the box lost 34.17188 J - 9.72 J = 24.451875 J of energy over a distance of 1.75 meters. Let's calculate the loss per meter by dividing the loss by the distance.

24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

Rounding to 1 decimal place gives 14.0 N which matches option "B".

8 0

2 years ago

Two objects each moving with speed v travel in opposite directions along a straight line passing through both their centers. The

Tpy6a [65]

Answer:

$\dfrac{1}{16}$

$\dfrac{5}{3}$

Explanation:

$m_1$ = Mass of first object

$m_2$ = Mass of second object

v = Speed of both objects

$\dfrac{v}{4}$ = Combined velocity

The ratio of final kinetic energy to initial kinetic energy will be

$\dfrac{K_f}{K_i}=\dfrac{\dfrac{1}{2}(m_1+m_2)(\dfrac{v}{4})^2}{\dfrac{1}{2}(m_1v^2+m_2v^2)} \\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{v^2}{16}}{m_1v^2+m_2v^2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{1}{16}}{m_1+m_2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{1}{16}$

The ratio is $\dfrac{1}{16}$

As the linear momentum is conserved

$m_1v-m_2v=(m_1+m_2)\dfrac{v}{4}\\\Rightarrow m_1-m_2=(m_1+m_2)\dfrac{1}{4}$

Divide by $m_2$ on both sides

$\dfrac{m_1}{m_2}-1=\dfrac{m_1}{4m_2}+\dfrac{1}{4}\\\Rightarrow \dfrac{m_1}{m_2}-\dfrac{m_1}{4m_2}=\dfrac{1}{4}+1\\\Rightarrow \dfrac{3m_1}{4m_2}=\dfrac{5}{4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5\times 4}{3\times 4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{3}$

The ratio of mass is $\dfrac{5}{3}$

6 0

3 years ago

An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust

zubka84 [21]

Answer:

Time to Reach Saturation = 0.0146 day

Explanation:

In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:

Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³

Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³

Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³

Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

Time to Reach Saturation = (50 mg)/(2.38 mg/min)

Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)

<u>Time to Reach Saturation = 0.0146 day</u>

3 0

3 years ago

What happens when the temperature of an object increases

tamaranim1 [39]

the average KE of the object increases

5 0

3 years ago