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Bond [772]
3 years ago
13

The mass of cart 1 was 1.10kg and mass of cart 2 is what's being solved for. They were next to each other and a spring was activ

ated and the two carts moved away from each other and produced these readings. Cart 1= v1: .75 m/s and Cart 2= v2: .34m/s. What's the mass of cart 2
Physics
1 answer:
klasskru [66]3 years ago
6 0
M1v1 + m2v2 = m1v1i + m2v2i 
fill it in m1 = mass 1 
m2 = mass 2
v1 = velocity 1
v2 = velocity 2 
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stepladder [879]

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A.   .33m

Explanation:

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Which parameter best defines the primary difference between weather and climate?
LenKa [72]

Answer:

time

Explanation:

weather is the atmospheric condition of a place over a short period of time, while climate is the weather condition prevailing in an area over a long period of time. From the two definitions above we can see that weather is the condition over a short period of time while climate is over longer periods, therefore the primary difference between them is time.      

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Which factor should change when comparing the boiling times of the liquids?
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7 0
3 years ago
Read 2 more answers
A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What
Oksanka [162]

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}

We know that,

m_{p}c^2=m_{np}c^2

So, E_{photon}=2mc^2+K.E_{p}+K.E_{np}

K.E_{np}=E_{photon}-(2mc^2+K.E_{p})

Put the value into the formula

K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}

K.E_{np}=147.4\ MeV

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0
3 years ago
A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Keith_Richards [23]

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}

I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}

I =0.032\ kg.m^2

using conservation of energy

\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}

\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}

\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}

\omega=2.71 \rad/s

we know,

v = r ω

v = \dfrac{L}{2} \times 2.71

v = \dfrac{0.8}{2} \times 2.71

v = 1.084 m/s

3 0
3 years ago
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