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OLEGan [10]
2 years ago
6

Lexi is balancing equations. She is finding one equation to be very difficult to balance. Which explains how to balance the equa

tion ZnSO4 + Li2CO3 → ZnCO3 + Li2SO4? *
1 point
Physics
1 answer:
Rina8888 [55]2 years ago
6 0

Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer

Equation is already balanced.

Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4

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Answer:

friction

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3 years ago
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

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2 years ago
Describe the forces on a rope that supports a tire swing when a child sits on the tire
noname [10]
Potential energy perhaps
7 0
2 years ago
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What does the zeroth law of thermodynamics allow us to define?
NeTakaya
It defines that if two thermodynamic systems are each in equilibrium with a third system, then they are in equilibrium with each other.
3 0
3 years ago
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A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
3 years ago
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