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Naya [18.7K]
3 years ago
10

Please HELP me, this is physics

Physics
1 answer:
dalvyx [7]3 years ago
3 0

Answer:

number 5 is 15 m

Explanation:

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Communicating results allows scientists to learn from each other's results.
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3 years ago
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A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?
sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

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

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

T=\frac{9.24}{15}=0.62 s

Now we can re-arrange the first equation  to solve for the mass:

m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg

Learn more about period:

brainly.com/question/5438962

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3 0
3 years ago
What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

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2 years ago
When the temperature of air rises, the amount of water needed for saturation
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It Increases. I just took a quiz with the same question.
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3 years ago
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