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RideAnS [48]
2 years ago
12

How do you break apart a

Physics
2 answers:
Tcecarenko [31]2 years ago
3 0

Answer:

B

Explanation:

Depends Mostly on bonds electrolysis can be used, chemical bonding like additional of water or by heating back to their elements.

nataly862011 [7]2 years ago
3 0
A.
Compounds are made chemically by two pure substances whereas mixtures can be only separated physically
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A car with mass 450 kg has a kinetic energy of 16,256 j. What is the speed of the car?
Likurg_2 [28]
E = 1/2 * m * v²

v = √(2 * E  / m)
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3 years ago
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Identify the areas on the image where the force of repulsion is the least.
ArbitrLikvidat [17]

Each magnet has a north pole and a south pole. We know that, from having played with bar magnets in our childhood, that a magnet's north pole will repel another magnet's north pole and attract its south pole.

From this diagram it is easy to see that the two lower bar magnets not only repel each other, but they are quite attracted to each other since their north and south poles are close together.

Therefore the region between the lower two magnets has the least force of repulsion.

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3 years ago
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What is the prefix for oxygen in As2O5? tri- di- mono- penta-
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3 years ago
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A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
3 years ago
Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?
Leto [7]
The answer is 19.3 j
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