What distance is required for a train to stop when traveling 50 miles per hour?

Identify the elements that come before iron (Fe) in the periodic table Fe, Co, Cu, K, Ni, Mn

harkovskaia [24]

FE is iron CO is cobalt CU is copper K is potassium NI is nickle MN is magnemese

5 0

3 years ago

The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 rpm. These

alexdok [17]

Answer:

θ = 6.3 *10³ revolutions

Explanation:

Angular acceleration of the drill

We apply the equations of circular motion uniformly accelerated

ωf= ω₀ + α*t Formula (1)

Where:

α : Angular acceleration (rad/s²)

ω₀ : Initial angular speed ( rad/s)

ωf : Final angular speed ( rad

t : time interval (s)

Data

ω₀ = 0

ωf = 350000 rpm = 350000 rev/min

1 rev = 2π rad

1 min= 60 s

ωf = 350000 rev/min =350000*(2π rad/60 s)

ωf = 36651.9 rad/s

t = 2.2 s

We replace data in the formula (2) :

ωf= ω₀ + α*t

36651.9 = 0 + α* (2.2)

α = 36651.9 / (2.2)

α = 17000 rad/s²

Revolutions made by the drill

We apply the equations of circular motion uniformly accelerated

ωf²= ω₀ ²+ 2α*θ Formula (2)

Where:

θ : Angle that the body has rotated in a given time interval (rad)

We replace data in the formula (2):

(ωf)²= ω₀²+ 2α*θ

(36651.9)²= (0)²+ 2( 17000 )*θ

θ = (36651.9)²/ (34000 )

θ = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)

θ = 6288.31 revolutions

θ = 6.3 *10³ revolutions

3 0

4 years ago

An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

3 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago

MULTIPLE CHOICE QUESTION

Nookie1986 [14]

Which object? More information is needed to answer this question

4 0

3 years ago