In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.
TRUE.
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
Answer:
Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters
Explanation:
First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).
Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.
The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.
Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)
Hello!
My best guess would be hydrogen and oxygen.
hopefully this helps!
Answer:

Explanation:
given,
Angular speed of the tire = 32 rad/s
Displacement of the wheel = 3.5 rev
Δ θ = 3.5 x 2 π
= 7 π rad
now,
Time interval of the car to rotate 7π rad
using equation



Time taken to rotate 3.5 times is equal to 0.687 s.