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RSB [31]
3 years ago
12

A startled armadillo leaps upward, rising 0.537 m in the first 0.220 s. (a) what is its initial speed as it leaves the ground? (

b) what is its speed at the height of 0.537 m? (c) how much higher does it go? use g=9.81 m/s2.
Physics
1 answer:
Elena L [17]3 years ago
3 0
Let u = initial vertical velocity.

Assume that
g = 9.81 m/s²,
Wind resistance is ignored.

When t = 0.220 s, the height is h = 0.537 m. Therefore
0.537 m = (u m/s)*(0.220 s) - (1/2)*(9.81 m/s²)*(0.220 s)²
0.537 = 0.22u - 0.2372
u = 3.519 m/s

The upward velocity after 0.220 s is
v = 3.519 - 9.81*0.22 = 1.363 m/s

At maximum height, the upward velocity is zero. The maximum height, H, is given by
(3.519 m/s)² - 2*(9.81 m/s²)*(H m) = 0
12.3834 - 19.6H = 0
H = 0.632 m
It goes higher by 0.632 - 0.537 = 0.095 m

Answers:
(a) The initial speed is 3.519 m/s.
(b) The speed at  0.537 m height is 1.363 m/s.
(c) It goes higher by 0.095 m.

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Answer:

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(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = 2\pi radian

(a) So angular speed =\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec

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Help with 2 Physics questions, WILL CHOOSE BRAINLIEST
Tju [1.3M]

1) D

2) D.) Greater than \theta_c

Explanation:

1)

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

  • The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
  • The angle of incidence is greater than a certain angle called critical angle

In picture 1, we have 4 different diagrams. In the diagrams:

  • The red arrow represents the incident ray
  • The green arrow represents the refracted ray
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Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

2)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

D.) Greater than \theta_c

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Answer:

The answer is at x = 0, which represents position B

Explanation:

The full question is:

"A block is attached to a horizontal spring and set in a

simple harmonic motion, as shown from above in the figure. When the spring is relaxed, the block is a position B, where the displacement x from the equilibrium position is 0. Letting D represent the maximum displacement, the extremes of the block's motion are at position A, where x= -D, and at position C, where x= D.

At what point in the motion is the speed of the block at its maximum?"

And you can see the figure on the attached file.

Simple Harmonic motion equations

We can start from the equation that describes the position that is

x(t)=D \sin\left(\omega t)

Here D stands for the amplitude which is the maximum displacement, and \omega is the angular velocity, thus we can find the derivative to find the velocity equation, so we get

v(t)=D \omega \cos (\omega t)

And we can find the derivative again to find the acceleration.

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We reach the maximum speed when the acceleration equation is equal to 0,

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Thus it happens when

\sin (\omega t)=0

So if we replace that on the position equation we get

x(t)=D \sin(\omega t) \\x(t)=D(0)\\x(t)=0

Thus the position where the speed of the block is at at its maximum is when it is going back to the origin, that is x = 0, so point b.

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