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marta [7]
3 years ago
12

Select the correct answer. x y 2.5 6.25 9.4 88.36 15.6 243.63 19.5 380.25 25.8 665.64 The table lists the values for two paramet

ers, x and y, of an experiment. What is the approximate value of y for x = 4? A. 11 B. 16 C. 24 D. 43
Physics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer: B. 16

Explanation:

The given table:

x y

2.5 6.25

9.4 88.36

15.6 243.63

19.5 380.25

25.8 665.64

Here, we observed that value of y is the square value of x.

i.e. y=x^2  [For example 2.5²=6.25, 9.4²=88.36, 15.6²=243.63]

Put x=4, we get

y=4^2=16

Hence, the approximate value of y for x = 4 is 16.

So, the correct option is B. 16 .

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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

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Here,

m = Mass

v = Velocity

k = Spring constant

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Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

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v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

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3 years ago
An airplane accelerates down a runmway at 2.9 m/s/s for 38.5 seconds
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Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

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s=1/2(3.20)(32.8)²

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3 years ago
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What is a normal force?
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Is the component perpendicular to the surface on contact  of the  contact force  <span />
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two children sit on a merry go round. on sits 0.5 m from the center and the other sits 3 m from the center. Which child has grea
GuDViN [60]

i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.

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3 years ago
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang
valina [46]

Answer:

work done is -2.8  × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta

W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta

W = -mgl  \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta

W = -mgl[ -cosθ ]^{0.047}_{0.045 }

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8  × 10⁻⁶ J

Therefore, work done is -2.8  × 10⁻⁶ J

6 0
3 years ago
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