A Stegosaurus hits' a wall with a force of

Physics

1 answer:

Pavel [41]3 years ago

3 0

Answer: a= -1.9 m/s²

Explanation: F = ma and a = F / m = 60 N / 32 kg = 1.875 m/s²

Because of law of action and reaction, wall applies same force to dino

In opposite direction and Dino is decelerating. Thus sign Is negative.

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Why the object will not return to its original position? why does the centre of mass fall?

dusya [7]

Answer:

the object will not return to the original position because it will not have any forces helping it to go up the hill

the center of mass falls since it is going downwards

Hope this helps!! have a great day

Explanation:

8 0

4 years ago

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

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

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:

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

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