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:
a. by moving the book without acceleration and keeping the height of the book constant
Explanation:
FOR CONSTANT KINETIC ENERGY:
The kinetic energy of a body depends upon its speed according to its formula:
ΔK.E = (1/2)mΔv²
So, for Δv = 0 m/s
ΔK.E = 0 J
So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.
FOR CONSTANT POTENTIAL ENERGY:
The potential energy of a body depends upon its height according to its formula:
ΔP.E = mgΔh
So, for Δh = 0 m/s
ΔP.E = 0 J
So, for keeping potential energy constant, the books must be moved at constant height.
So, the correct option is:
<u>a. by moving the book without acceleration and keeping the height of the book constant</u>
Answer:
The appropriate solution is "2.78 mm".
Explanation:
Given:

or,



or,

As we know,
Fringe width is:
⇒ 
hence,
Separation between second and third bright fringes will be:
⇒ 


or,

Answer:
Q = 200800 Joules.
Explanation:
Given the following data;
Mass = 4kg
Initial temperature = 30.0°C
Final temperature = 90.0°C
Specific heat capacity of glass = 837 J/kg°C
To find the quantity of heat absorbed;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt = T2 - T1
dt = 90 - 30
dt = 60°C
Substituting the values into the equation, we have;
Q = 200800 Joules.
Therefore, the amount of heat absorbed is 200800 Joules.
Initially, the velocity vector is
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get 