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) 0.568 kg
b) 474 kg/m³
Explanation:
Given:
Inner radius = 8.82 cm = 0.0882 m
Outer radius = 9.91 cm = 0.0991 m
Density of the liquid = 948.00 Kg/m³
a) The volume of the sphere = 
or
volume of sphere = 0.0012 m³
also, volume of half sphere = 
or
volume of half sphere = 
or
Volume of half sphere =0.0006 m³
Now, from the Archimedes principle
Mass of the sphere = Weight of the volume of object submerged
or
Mass of the sphere = 0.0006× 948.00 = 0.568 kg
b) Now, density = 
or
Density = 
or
Density = 474 kg/m³
Answer:
Explanation:
The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:
Where:
Asumming v1=0 and t1=0:
To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.
In the case of work, we know that it is defined by,
Where,
F= Force
d = Distance
The distance in this case is a composition between number of steps and the height. Then,
, for h as the height of each step and N number of steps.
On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)
Where,
Final velocity
Initial Velocity
a = Acceleration
Displacement
PART A) For the particular case of work we know then that,
Therefore the Work to do that activity is 4.41kJ
PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,
Here,
3 steps in one second
Replacing,
Re-arrange for a,
At this point we can calculate the time, which is,
With time and work we can finally calculate the power
P = \frac{W}{t} = \frac{4.41}{20}
P = 0.2205kW
Answer:
The answer is "a, c and b"
Explanation:
- Its total block power is equal to the amount of potential energy and kinetic energy.
- Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
- Because the block in the first case has no initial speed, the block has zero film energy.
- For both the second example, it also has the
velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher. - While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
- The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.
