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Slav-nsk [51]
3 years ago
15

The two ways you could increase acceleration

Physics
1 answer:
GalinKa [24]3 years ago
5 0

Answer: mass(how much the object weighs) and force(how hard you push the object)

Explanation:

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Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

\epsilon_{max} = NBA\omega

Where,

N= Number of turns of the coil

B = Magnetic field

\omega = Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

\omega = 2\pi f

\omega = 2\pi*61.5

\omega =123\pi rad/s

Replacing at the equation our values given we have that

\epsilon_{max} = NBA\omega

\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega

\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)

\epsilon_{max} = 1.5732*10^{-11}V

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

6 0
4 years ago
A stone is dropped down a well and hits the water 2.50 s later. What is the depth from the edge of the well to the water? 61.3,
Gennadij [26K]

Answer;

30.6 m

Explanation;

All objects accelerate at the constant rate in the Earth's gravitational field. The gravitational acceleration, g = 9.8 m/s².

Distance traveled by an object falling down under a constant acceleration will be given the formula;

s = ut² + 1/2(gt²); but u the initial velocity is o

thus;

S =1/2(gt²)

  = 0.5 × 9.81 × 2.5 ²

  = 30.65

  ≈ 30.6 m

5 0
3 years ago
As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north
dedylja [7]
<span>in this case velocity of airoplane will be 750*sin20'


</span>
7 0
4 years ago
Read 2 more answers
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
4 years ago
A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele
kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh'            

h'=\dfrac{v^2}{2g}

h'=\dfrac{(10)^2}{2\times 9.8}

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.                              

6 0
3 years ago
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