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2 years ago

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A spinning top completes 6.00x10^3 rotations before it starts to topple over. The average speed of the rotations is 8.00x10^2 rp

m. Calculate how long the top spins before it begins to topple

1 answer:

Marianna [84]2 years ago

4 0

Answer:

Explanation:

This is simple logic. If the top can spin 6000 times before it starts to topple, and each minute the top can spin 800 times,

6000/800 = 7.50 minutes

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Speakers A and B are vibrating in phase. They are directly facing each other, are 8.2 m apart, and are each playing a 78.0 Hz to

Stels [109]

Answer:6.298,4.1,1.9015

Explanation:

Wavelength= $\frac{velocity of sound }{frequency}$

= $\frac{343}{78}$ =4.397 m

Distance of 3rd speaker from speaker A is x

From B 78-x

Difference between the distances must be a whole number of wavelengths

First

$x-\left ( 8.2-x\right )=4.397$ for 1 st wavelength

2x=8.2+4.397=12.597

x=6.298m

second

For zero wavelength

$x-\left ( 8.2-x\right )=0$

2x=8.2

x=4.1m

Third

$\left ( 8.2-x\right )-x=4.397$

x=1.9015 m

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Volcanic ash and sulfur dioxide spewed out of Mt. Pinatubo in 1991. These materials can reflect incoming solar radiation. Over t

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3 years ago

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A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

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3 years ago

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3 years ago

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Lol i'm going to fail please help

Novosadov [1.4K]

Answer:

Explanation:

The frequency is 16.0 Hz. That means that 16 of these waves can pass a single point in 1 second. We are given frequency and wavelength. The equation that relates them is

$f=\frac{v}{\lambda}$ where f is frequency, v is velocity, and λ is wavelength. Putting all this together:

$16.0=\frac{v}{97.5}$ and solving for velocity,

v = 16.0(97.5) so

v = 1560 m/s. This wave can travel 1560 meters in a single second!!! Now that we know this velocity, we can use it in a proportion to find our unknown, which is how long, t, it will take to hear this sound 11000m away. (11 km is 11000m):

$\frac{1560m}{1s}=\frac{11000}{t}$ and cross multiply to get

1560t = 11000 so

t = 7.1 seconds

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3 years ago