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3 years ago

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table.

Physics

2 answers:

musickatia [10]3 years ago

8 0

Answer:

B.) X,W,Y,Z

Explanation:

The elastic potential energy equation is:

PE = (1/2)*k*x²

where x is distance and k is the spring constant. Then, as k increases, PE also increases, so, the greatest potential energy corresponds to the greatest k, and vice versa.

Natalka [10]3 years ago

4 0

The Hook's law states that F=Ke, where
F⇒force applied to the spring,
e⇒extension of the spring and
K⇒spring constant.
Spring constant determines the strength of the spring. The larger the spring constant the stronger the spring it is.
The list from the greatest to list is:

X, W, Y, Z

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3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

Brut [27]

Answer:

$\boxed{\sf Work \ done = 4 \ J}$

Given:

Force = 8 N

Distance covered by the body = 50 cm = 0.5 m

Explanation:

Work Done = Force × Distance covered by the body

= 8 × 0.5

= 4 J

6 0

3 years ago

A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

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With this function we should only calculate the derivate in function of c

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That is the rate of change of $\theta$ .

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Hence we have finally the rate of change when c=0.

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3 years ago

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Alex Ar [27]

I'm quite sure it is A

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4 years ago

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