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.
If a man has a mass of 83 kilograms on Earth, the force of gravity on his body be on the moon 135.6N. force =mass*acc , 83 * 9.8/6= 813.4/6 = 135.6N
Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m
