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mamaluj [8]
3 years ago
5

Identify the basic structure of a chemical equation

Physics
1 answer:
guajiro [1.7K]3 years ago
6 0
The basic structure would be:

Reactants → Products 
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Sound source A is located at x = 0, y = 0, and sound source B is located at x = 0, y = 2.3 m. The two sources radiate coherently
Oliga [24]

Answer:

Solution is given in the attachments

4 0
3 years ago
What two things are necessary for work to be done on an object?
NemiM [27]

Answer: Force and Movement

Explanation:

The first is that the object moves. The second is that a force must act on the object in the direction the object moves.

6 0
3 years ago
A 3.1-kilogram gun initially at rest is free to
34kurt
The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.
8 0
3 years ago
Read 2 more answers
Which has more volume feathers or rocks
Nataly_w [17]
Volume doesn't depend on what the substance is, only on how much of it there is.
5 0
3 years ago
A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
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