Answer:
0.932 m/h
Explanation:
Parameters given:
Distance walked by couple, D = 1.5 km
Time spent in walking, t = 60 min
To solve this we simply apply the formula for average peed:
Average Speed = total distance / total time taken
The question asks specifically to put the answer in miles per hour (m/h), hence, we need to convert the distance to miles and the time to hours:
Distance in miles:
1 km = 0.622 miles
1.5 km = 1.5 * 0.622 = 0.932 miles
Time in hours = 60 / 60 = 1 hour
Hence, the speed is:
Speed = 0.932 / 1 = 0.932 m/h
The average speed of the couple is 0.932 m/h
If it happens to be cruising along at 5 meters per second (about
11 miles per hour), then it covers 10 meters in 2 seconds without
any acceleration at all.
Answer:
B. Vestibular nuclei
Explanation:
The nerve information generated by the vestibular receptors travels through the vestibular portion of the eighth pair that penetrates the brain stem at the level of the brain stem bridge. At this level there are four vestibular nuclei, which receive the synapses of these axons, coming from the ridges and macules. The semicircular ducts predominantly terminate in the superior and medial nuclei. While the fibers coming from the macules end on the lateral, medial and inferior nuclei. Some fibers of the eighth pair end in the flocculonodular lobe of the cerebellum, <u>these connections play an important role in controlling posture and balance.</u>
From the vestibular nuclei, two bundles of fibers descending to the spinal cord originate from the medial and lateral vestibular spinal bundles and a bundle of fiber that rises in the brain stem that participates in the coordination of eye movements, the medial longitudinal fascicle, which participates in Rotational nystagmus This system also participates in an important way in the control of some ocular movements by the fibers that it contributes to the medial longitudinal fascicle, which is a structure that interconnects the motor nuclei of the extrinsic muscles of the eyeballs VI or abdicens nucleus (abductor) on one side and IV or pathetic nucleus (trochlear) and III or nucleus of the common ocular motor (oculomotor) on the opposite side.
Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N
<u>Explanation:</u>
<u />
Given:
Mass of the ball, m = 2kg
Density of ball, σ = 800 kg/m³
Density of water, ρ = 1000 kg/m³
Downward force acting by the ball in the vessel = mg
where, g = 9.8m/s²
F = 2 X 9.8
F = 19.6N
Net force acting on the ball:
Fnet = (ρ - σ) Vg
where,
V is the volume of water
Fnet = (1000 - 800) V X 9.8
Fnet = 1960V N
If the volume is known, then substitute the value of V to find the net force.
Thus, Downward force acting on the ball is 19.6N
Net force acting on the ball is 1960V N
The answer is ....... none of the above. The reactivity of an element is based on its valence electrons
