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4vir4ik [10]
3 years ago
7

Determine the stopping location of the prize wheel. At this moment it is centered on the number 11. It is spinning at a rate of

124.40 rpm. It is slowing at a rate of 1.400 rad/s/s.
Predict the time the wheel will remain spinning, the angular displacement it will go through, and then use this to predict the number on which the wheel will stop. To be safe, you will also get one number the right and to the left of the number you chose.
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer:

The wheel spins for 11.43s. The number cannot be determined.

Explanation:

We know that initial velocity is 124.4 rpm = 13.027137522 rad/s, acceleration is -1.14 rad/s^2, and final velocity is assumed to be 0 rad/s. We are asked to find t and displacement. We use the equation \omega=\omega_0+\alpha t where \omega=0 rad/s, \omega_0=13.027 rad/s, \alpha=-1.14rad/s^2, and t=?.

Rearrange the equation to obtain \frac{\omega-\omega_0}{\alpha}=t. Plug in the numbers and solve to obtain t=11.43s.

The number of the wheel cannot be determined as we do not know the placement of numbers on the wheel.

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Need a little help here :(
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Answer:

The output out be 200

Explanation:

Hope this helps :))

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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

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

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

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,

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

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}

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7 0
3 years ago
The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.1
mr_godi [17]

Answer:

Minimum number of photons required is 1.35 x 10⁵

Explanation:

Given:

Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m

Energy of one photon is given by the relation :

E=\frac{hc}{\lambda}    ....(1)

Here h is Planck's constant and c is speed of light.

Let N be the minimum number of photons needed for triggering receptor.

Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J

According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,

E₁ = N x E

Put equation (1) in the above equation.

E_{1}=N\times\frac{hc}{\lambda}

Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.

3.15\times10^{-14} =N\times\frac{6.6\times10^{-34}\times3\times10^{8}}{850\times10^{-9}}

N = 1.35 x 10⁵

8 0
3 years ago
If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, its t
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E_{total} = KE + PE \\ KE = 500 J \\ PE = 1500 J \\ E_{total} = 500J + 1500J \\ E_{total} = 2000J
4 0
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