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2 years ago

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A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

person traveling with the basket? show your work HINT: Use the law of conservation of momentum in the perfectly inelastic collision m1v1i + m2v2i = (m1+m2) vf
plz guys i have exam now answer her quikly

1 answer:

11Alexandr11 [23.1K]2 years ago

4 0

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

<u>v_f = 10.85 m/s</u>

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forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

$E_{kin}=E_{pot}$

Donde:

$E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h$

Ahora reemplazando:

$\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} } \\\\v=5.24[m/s]$

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3 years ago

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

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3 years ago

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The process of <u>scientific method</u> involves making hypotheses , driving predictions from them as logical consequences , and then carrying out experiments or empirical observations based on those predictions. A hypotheses is a conjecture , based on knowledge obtained while seeking answers to the questions.

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2 years ago

Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con

Andrew [12]

Answer:

$T= 2\pi\times sqrt(m/(k1+k2))$

Explanation:

When the block is displaced by x units

F= spring force

two springs are connected parallel

$F =-k_1x - k_2x$

Writing Newtons second law, F = ma

$-k_1x - k_2x =ma$

$-k_1x - k_2x = mx''$

a= x" ( differentiating x w.r.t time twice)

$x''+(k_1/m + k_2/m) x=0$

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

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2 years ago

two soccer players run toward each other. one player has a mass of 85 kg and runs west with a speed of 8 m/s, while the other ha

marissa [1.9K]

Answer:

Total momentum, p = 55 kg-m/s

Explanation:

It is given that,

Mass of player 1, m₁ = 85 kg

Mass of player 2, m₂ = 105 kg

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Speed of player 2, v₂ = 7 m/s (east)

Momentum is equal to the product of mass and velocity. For this system, momentum is given by :

$p=m_1v_1+m_2v_2$

$p=85\ kg\times (-8\ m/s)+105\ kg\times 7\ m/s$

p = 55 kg-m/s

The total momentum of the system made up of the two players is 55 kg-m/s.

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2 years ago

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