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icang [17]
3 years ago
10

A 350-g air track cart is traveling at 1.25 m/s and a 280-g cart traveling in the opposite direction at 1.33 m/s. What is the sp

eed of the center of mass of the two carts?
Physics
1 answer:
kumpel [21]3 years ago
4 0

Answer:

The speed of the center of mass of the two carts is 0.103 m/s

Explanation:

It is given that,

Mass of the air track cart, m₁ = 350 g = 0.35 kg

Velocity of air track cart, v₁ = 1.25 m/s

Mass of cart, m₂ = 280 g = 0.28 kg

Velocity of cart, v₂ = -1.33 m/s (it is travelling in opposite direction)

We need to find the speed of the center of mass of the two carts. It is given by the following relation as :

v_{cm}=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}

v_{cm}=\dfrac{0.35\ kg\times 1.25\ m/s+0.28\ kg\times (-1.33\ m/s)}{0.35\ kg+0.28\ kg}

v_{cm}=0.103\ m/s

Hence, this is the required solution.

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Answer:

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velocity - the rate of displacement of a moving object over time

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Explanation:

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Imagine you have just witnessed a small avalanche on a mountain while skiing, and two slushy snowballs just crashed together in
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We have that the momentum p is given by the formula p=mv where m is the mass and v is the velocity. Since for A p=-14kgm/s and m=7, we have that the velocity is -14/7=-2m/s. Hence its speed is 2 m/s.
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What is system thinking ?
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3 years ago
A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.
ohaa [14]

Answer:

speed of a particle  = 2.31 × 10^{6} m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = \frac{E}{B}    ...................3

put here value

V =  \frac{0.51*10^6}{0.22}

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and

now energy of proton required will be here as

energy of proton required = mass of proton × \frac{V^2}{2}

put here value

energy of proton required = 1.67 × 10^{-27} × \frac{(2.31*10^6)^2}{2}

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energy of proton required = 27777.777 eV

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and

now we get here energy of electron required that is

energy of electron required = mass of electron × \frac{V^2}{2}

put here value

energy of electron required = 9.11 × 10^{-31} × \frac{(2.31*10^6)^2}{2}

energy of electron required =  24.305× 10^{-19} J

energy of electron required =  24.305 × 10^{-19} J ÷ (1.602 × 10^{-19}  

energy of electron required =  15.171 eV

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