How many electrons in Cu

An angry rhino with a mass of 2600 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distr

worty [1.4K]

Answer:

vf₂ = 15.79 m/s

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

where

p:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ :Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁ = 2600 kg : mass of the rhino

m₂= 0.18 kg : mass of the ball

v₀₁ = 3.70 m/s : initial velocity of the rhino

v₀₂= - 8.39 m/s, initial velocity of the ball

Problem development

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

(2600)*(3.7) + (0.18)*(- 8.39) = (2600)*vf₁ +(0.18)*vf₂

9620 -1.5102 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

$e = \frac{v_{f2}-v_{f1} }{v_{o1}-v_{o2}}$

1*( v₀₁- v₀₂ ) = (vf₂ -vf₁)

( 3.7 -( -8.39 ) = (vf₂ -vf₁)

12.09 = (vf₂ -vf₁)

vf₂ = vf₁ + 12.09 Equation (2)

We replace Equation (2) in the Equation (1)

9618.4898 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)* (vf₁ + 12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + (0.18)(12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + 2.1762

9618.4898 -2.1762 = (2600.18)*vf₁

9616.3136 = (2600.18)*vf₁

vf₁ = (9616.3136) / (2600.18)

vf₁ = 3.698 m/s : Final velocity of the rhino

We replace vf₁ = 3.698 m/s in the Equation (2)

vf₂ = vf₁ + 12.09

vf₂ = 3.698 + 12.09

vf₂ = 15.79 m/s : Final velocity of the ball

6 0

3 years ago

Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration

stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is $\mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}$

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

$F=G\frac{Mm}{r^{2}}$ (1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

$M=5.972\times10^{24}\,kg$

$m=7.34767309\times10^{22}\,kg$

$r=384400\,km$

$G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}$

Using those values on (1)

$F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}$

$F\approx1.98193\times10^{20}N$

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

$F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}$

6 0

4 years ago

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

yawa3891 [41]

Answer:

Isothermal : P2 = ( P1V1 / V2 ) , work-done $pdv = nRT * In( \frac{V2}{v1} )$

Adiabatic : : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$ , work-done =

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

$pdv = nRT * In( \frac{V2}{v1} )$

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

$P1V1^y = P2V2^y$

where y = 5/3

hence : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$

Work-done

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Where $T2 = T1V1^(2/3)/V2^(2/3)$

3 0

3 years ago

Hi guys . I'm having some troubles with physics and chemistry .

Otrada [13]

Answer:

physics and math tutor

Save my exams

Explanation:

hii! for PTM(physics and math tutor) you simply type in the search bar for example in. physics Units and measurements past papers.Then the first option would be PTM(physics and math tutor) and you click on it and it would give you on the left side Revision Notes, flashcards and practice questions. This site has helped me so much and I hope this helps you too (by the way PTM has for all 3 sciences and math too) just simply type the lesson name along with past paper and click on the first link and you'll get yourself your answers!!!! hopefully this helps!:))

6 0

3 years ago

Describe using examples how objects can be at rest and in motion simultaneously

elena-14-01-66 [18.8K]

An object can be at rest and still be in motion because the earth is always in motion.

5 0

3 years ago

How many electrons in Cu​

How many electrons in Cu