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Andreas93 [3]
3 years ago
9

Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon'

s tongue accelerates at a remarkable 210 m/s^2 for 20 ms, then travels at constant speed for another 30 ms.
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
icang [17]3 years ago
6 0

Answer:

  x_total = 0.17m

Explanation:

We can treat this exercise with the kinematics equations, where in the first part it is accelerated and in the second it is a uniform movement.

Let's analyze accelerated motion

The time that lasts is t = 20 10⁻³ s, the initial speed is zero (v₀ = 0), let's find the length that advances

            x₁ = v₀ t + ½ a t²

            x₁ = ½ a t²

            x₁ = ½ 210 (20 10⁻³)²

            x₁ = 4.2 10⁻² m

 

let's find the speed for the end of this movement

            v = v₀ + a t

            v = 0 + 210   20 10⁻³

            v = 4.2 m / s

with this speed we can find the distance that the uniform movement

           x₂ = v t2

           x₂ = 4.2   30 10⁻³

           x₂ = 1.26 10⁻¹ m

           x₂ = 0.126m

the total distance traveled is

          x_total = x₁ + x₂

          x_total = 0.0420 +0.126

          x_total = 0.168m

           

     Let's reduce the significant figures to two

          x_total = 0.17m

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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

where:

\omega=6.28 rad/s is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

v=(6.28)(0.65)=4.1 m/s

Learn more about angular speed:

brainly.com/question/9575487

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brainly.com/question/2506028

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7 0
3 years ago
A compound microscope is equipped with two objective lenses (10x and 45x) and has a 10x ocular lens. The highest magnification a
Wittaler [7]

Answer:

450X

Explanation:

When a specimen is been viewed, both

objective and ocular lenses works together so that the object is magnified.

From the question,objective lenses are;

1)10x

2)45x

ocular lens= 10x

Highest magnification

= 10X ocular × 45X objective

=450X

This implies that the image that was viewed will appear 450 times the actual size.

5 0
2 years ago
A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped
Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

6 0
2 years ago
A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0
3 years ago
(please help i gotta turn this in a few minutes 10 points!)
pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

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7 0
3 years ago
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