Answer:
The evolutionary success of bats is accredited to their ability, as the only mammals, to fly and navigate in darkness by echolocation, thus filling a niche exploited by few other predators. Over 90% of all bat species use echolocation to localize obstacles in their environment by comparing their own high frequency sound pulses with returning echoes. The ability to localize and identify objects without the use of vision allows bats to forage for airborne nocturnal insects, but also for a diverse range of other food types including motionless perched prey or non-animal food items.
The agility and precision with which bats navigate and forage in total darkness, is in large part due to the accuracy and flexibility of their echolocation system. The echolocation clicks of the few echolocating Pteropodidae (Rousettus) are fundamentally different from the echolocation sounds produced in the larynx that we focus on here, and thus not part of this review. Many studies have shown that bats adapt their echolocation calls to a variety of conditions, changing duration and bandwidth of each call and the rate at which calls are emitted in response to changing perceptual demands . In recent years the intensity and directionality of echolocation signals has received increasing research attention and it is becoming evident that these parameters also play a major role in how bats successfully navigate and forage. To perceive an object in its surroundings, a bat must ensonify the object with enough energy to return an audible echo. Hence, the intensity and duration of the emitted signal act together to determine how far away a bat can echolocate an object. Equally important is signal directionality. Bat echolocation calls are directional, i.e., more call energy is focused in the forward direction than to the sides (Simmons, 1969; Shimozawa et al., 1974; Mogensen and Møhl, 1979; Hartley and Suthers, 1987, 1989; Henze and O'Neill, 1991). An object detectable at 2 m directly in front of the bat may not be detected if it is located at the same distance but off to the side. Consequently, at any given echolocation frequency and duration, it is the combination of signal intensity and signal directionality that defines the search volume, i.e., the volume in space where the bat can detect an object.
The aim of this review is to summarize current knowledge about intensity and directionality of bat echolocation calls, and show how both are adapted to habitat and behavioral context. Finally, we discuss the importance of active motor-control to dynamically adjust both signal intensity and directionality to solve the different tasks faced by echolocating bats.
Explanation:
Answer:
The time rate of change of flux is 
Explanation:
Given :
Current 
A
Area of plate 
Plate separation 
m
(A)
First find the capacitance of capacitor,
Where 
F
But 
Where 
Now differentiate above equation wrt. time,
Therefore, the time rate of change of flux is
The cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
Answer:
Explanation:
Acceleration exerted by an object is the measure of change in speed or velocity of that object with respect to time. So the initial and final velocities play a major role in determining the acceleration of the cyclist. As here the initial velocity of the cyclist is the speed at rest and that is given as 0 m/s. Then after 3 seconds, the velocity of the cyclist changes to 9 m/s.
Then acceleration = change in velocity/Time.
Acceleration = (9-0)/3=9/3=3 m/s².
So the cyclist accelerates from 0 m/s to 9 m/s in 3 seconds with an acceleration of 3 m/s².
1. B- it can cause cirrhosis
2. C- intoxication
3. A- admit to having a problem with alcohol
4. D- binge-drinking
5. A- depressant
6. true
7. false
Answer:
The heat transferred through the wall that day is 13728 BTUs
Explanation:
Here, we have the area of the wall given as
Area of wall = 2 × Length × Height + 2 × Width × Height
Length = 15 feet
Width = 11 Feet and
Height = 9 feet
Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²
Temperature difference is given by
Average outside temperature - Wall temperature = 40 - 18 = 22 °F
Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;
468 × 22 × 24/18 = 13728 = 13728 BTUs.
