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3 years ago

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The tachometer in the author's car has a maximum reading of 16,000 rpm. (A tachometer is a gauge that shows the angular speed of

a car's engine.) Express this angular speed in rad/s.

1 answer:

Liono4ka [1.6K]3 years ago

8 0

Answer:

the angular speed in rad /s is 1675.516 rad/s

Explanation:

The computation is shown below;

The Tachometer's reading is

= 16000 rpm

= 266.67 rps

Now

angular speed ,w = 2 × $\pi$ ×freq

= 2 × 3.14 × 266.67

= 1675.516 rad/s

Hence, the angular speed in rad /s is 1675.516 rad/s

The same is to be considered by applying the above formula so that the correct speed could arrive

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Where should clouds form and why

LiRa [457]

Answer:

When air rises in the atmosphere it gets cooler and is under less pressure. When air cools, it's not able to hold all of the water vapor it once was. Air also can't hold as much water when air pressure drops. The vapor becomes small water droplets or ice crystals and a cloud is formed.

Explanation:

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3 years ago

The element chlorine is found in nature as a molecule that can be represented as - A. Cl.

poizon [28]

A
Cl is the chemical symbol for chlorine numbers after it are isotopes

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3 years ago

A football player collides with another player. The force of the collision is

iragen [17]

Answer:

Impulse = 1000 Ns

Explanation:

Given the following data;

Force of collision = 1000 kg•m/s.

Time = 1 seconds

To find the impulse;

Mathematically, the impulse experienced by an object or body is given by the formula;

Impulse = force * time

Substituting into the formula, we have;

Impulse = 1000 * 1

Impulse = 1000 Ns

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3 years ago

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred

madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

Given that

T₁ = 300 K ,V₁ = 3 m³ ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas P V = m R T

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

$P_2=\dfrac{T_2}{T_1}\times P_1$

$P_2=\dfrac{600}{300}\times 2$

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

$m=\dfrac{PV}{RT}$

$m=\dfrac{200\times 3}{0.287\times 300}\ kg$

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0 Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

$S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}$

$S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}$

S₂ - S₁ = 3.42 KJ/K

5 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago