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Arisa [49]
3 years ago
12

A car sitting at a red light begins to accelerate at 2.0 m/s2 when the light turns green. It continues with this acceleration un

til it reaches a speed of 20 m/s. It then travels at this speed for another few minutes. How far does the car travel in the first 46 s after the light changes to green?
Physics
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

d = 820 m

Explanation:

here we know that car starts from rest and continue to accelerate till it will reach to maximum speed of 20 m/s

so we will have

v_f - v_i = at

20 - 0 = 2t

t = 10 s

so car will accelerate till t = 10 then it will move with uniform speed

so the distance moved by the car till it accelerates is given as

d = \frac{1}{2}at^2

d = \frac{1}{2}(2)(10^2)

d_1 = 100 m

now it will move with uniform speed for next 36 s

so we have

d_2 = 20(36) = 720 m

so total distance moved by the car is given as

d = d_1 + d_2

d = 100 + 720

d = 820 m

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Answer:

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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:

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3 years ago
Energy is the ability to
bekas [8.4K]
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3 years ago
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Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

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W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

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3 years ago
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stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

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Substituting in centripetal acceleration equation,

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Option D is the correct answer.                

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Answer:

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on edg2020

Explanation:

I got it right on the test

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