If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.
The values of the velocities are,
We know that the acceleration is equivalent to the change of the speed in a certain time therefore
Now applying the Newton's second law we have,
Therefore the approximate magnitude is 8516.36N
Answer:
15000 m/s
Explanation:
You just need to multiply the wavelength with the frequency.
Answer:
No, the magnitude of the magnetic field won't change.
Explanation:
The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.
