Newton's three laws of motion can be used to describe the motion of the ice skating.
<h3>Newton's first law of motion</h3>
Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
- Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.
<h3>Newton's second law of motion</h3>
Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.
- Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.
<h3>Newton's third law of motion</h3>
This law states that action and reaction are equal and opposite.
- Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.
Learn more about Newton's law here: brainly.com/question/3999427
The synapse is actually the link between 2 neurons. Now when
an action potential contacts the synaptic knob of a neuron, the voltage-gate
calcium channels are unlocked, resulting in an influx of positively charged
calcium ions into the cell. This makes the vesicles containing
neurotransmitters, for example acetylcholine, to travel towards the
pre-synaptic membrane. When the vesicle arrives at the membrane, the contents
are released into the synaptic cleft by exocytosis. Neurotransmitters disperse
across the space, down to its concentration gradient, up until it reaches the
post-synaptic membrane, where it connects to the correct neuroreceptors. Connecting
to the neuroreceptors results in depolarisation in the post-syanaptic neuron as
voltage-gated sodium channels are also opened, and the positively charged
sodium ions travel into the cell. When adequate neurotransmitters bind to
neuroreceptors, the post-synaptic membrane overcame the threshold level of
depolarisation and an action potential is made and the impulse is transmitted.
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
ripples on the surface of water.
vibrations in a guitar string.
a Mexican wave in a sports stadium.
electromagnetic waves – eg light waves, microwaves, radio waves.
seismic S-waves.
