Answer:
Volume of water displaced = 450 - 375 = 75 ml
Vr = volume of rock = 75 ml
Wr = 9.22 - 7.60 = 1.62 N weight of 75 ml of rock
Density of rock = 1.62 N / 75 ml = .0216 N / ml
Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml
Density of rock / density of water = .0216 / .0098 = 2.20
B. A hypothesis is just an idea. And can become a Theory if there is some evidence. However a Theory is not a law, but an idea with supportive evidence
Answer:
<em>The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.</em>
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

- where Rint = r and RL = r
- Replacing these values in I₁, we have:

- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

- We can find the relationship between I₂, and I₁, dividing both sides, as follows:

- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.