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Sidana [21]
3 years ago
6

A swing oscillates 8 times in 12 seconds what is its period what is its frequency

Physics
1 answer:
kotykmax [81]3 years ago
4 0
[Period (T)] T=time/# of cycles
T=12/8=3/2

[Frequency (F)] F=# of cycles/time
F=8/12=2/3

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Calculate the speed of a bus that travels a distance of 55 miles in 0.75 hours.
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Speed = distance / time = 55 / 0.75 = 73.3 miles per hour.
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3 years ago
Some bats have specially shaped noses that focus ultrasound echolocation pulses in the forward direction. Why is this useful?
creativ13 [48]

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:

3 0
3 years ago
Witch type of radio station can broadcast a signal the gratis distance
Nana76 [90]
AM radio stations I believe
6 0
3 years ago
The speed of an object and direction of its travel is?
makkiz [27]
Velocity is the answer..

hope that helps
5 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
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