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Allisa [31]
2 years ago
11

2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is

applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?

Physics
2 answers:
larisa86 [58]2 years ago
8 0

Explanation:

Below is an attachment containing the solution.

vodka [1.7K]2 years ago
4 0

Answer:

a) the distance moved by the object is 14m

b) the speed of the box at 3.00s is - 18.5m/s

The negative sign in the answer indicates that the speed of the box is in opposite direction. Therefore, the direction of the box is towards left.

Explanation:

see the attached file

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Answer:

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Answer:

5.444\times 10^{-4}

Explanation:

The momentum of the neutron before and after the decay  is the same since there's no external force.

P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

#The neutron is initially at rest, so after the decay:

P_A+P_B=0\\\\P_A=-P_B

#After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

Therefore:

P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

Let the energy released during the decay be Q:

Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

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4 1
2 years ago
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vlabodo [156]
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4 0
2 years ago
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Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
ratelena [41]

Answer:

the yield of product is YP=46.835 % and the concentration of solids is

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Explanation:

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then the yield YP of the product is

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Mass of solids in 100 kg of milk = 100 kg* 0.128 = 46.835 kg * Cs

Cs = 100 kg* 0.128 / 46.835 kg  = 0.2733 = 27.33%

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Cloud [144]

Answer:

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\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

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