C.) 65 km =<br> mm<br> Convert

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

$v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_{B2}-v_{A2}$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

$2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s$

7 0

4 years ago

A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s

Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

p₀ = m v₀

celestines que clash case

p_f = (m + M) v

po = pf

m v₀ = (n + M) v

v = $\frac{m}{m+M}$

calculemos

v= $\frac{0.200}{0.200+M} 0.450$

v= 0.09 m/s

elastic shock case

p₀ = m v₀

p_f = m v₁ +M v₂

p₀ = p_f

m v₀ = m v₁ + m v₂

6 0

3 years ago

Based on details in The Riddle of the Rosetta Stone, which statement best describes Jean-Baptiste Fourier's influence on Jean-Fr

Ira Lisetskai [31]

Answer:

hello your question is incomplete below is the missing part of the complete question

which statement best describes Jean-Baptiste Fourier’s influence on Jean-François Champollion? Fourier shared his research about mathematics with Champollion, which aided him in his research. Fourier was the first to introduce Champollion to the hieroglyphs on the Rosetta Stone, which fascinated him. Fourier shared his findings about the Rosetta Stone with Champollion, which helped him in his research. Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages.

answer : Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages.

Explanation:

Jean-Baptiste Fourier's influence on Jean-Francois Champollion is that Fourier was the first to introduce Champollion to a variety of languages, which prompted his love of languages. also Fourier was a mathematician who pleaded on the behalf of Champollion when he was about to be adopted into the military.

5 0

3 years ago

Read 2 more answers

If we model a blade as a point mass at the midpoint of the blade, what is the inward force necessary to provide each blade's cen

Alex17521 [72]

Answer:

F = 1.2x 10^6 N

Explanation:

The explanation is in the attached file