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Virty [35]
3 years ago
8

What is the period and group of gold

Physics
1 answer:
algol133 years ago
4 0

Answer:

Chemical elements, a dense lustrous yellow precious metal of group.

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A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a
Tom [10]

Answer:

The value of new value of angular speed of merry go round.\omega_{2} = 0.96 \frac{rad}{sec}

Explanation:

Given data

r = 1.4 m

Moment of inertia I_{1} = 265 kg - m^{2}

N_{1} = 11 RPM

\omega_{1} = \frac{2 \pi N}{60}

\omega_{1} = \frac{2 \pi (11)}{60}

\omega_{1} = 1.15 \frac{rad}{sec}

From conservation of momentum principal

I_{1} \omega_{1}  = I_{2} \omega_{2} ------- (1)

I_{2} = m r^{2} + 265

I_{2} = 27 (1.4)^{2} + 265

I_{2} = 317.92 \ kg m^{2}

Put all the values in equation  (1)

265 × 1.15 = 317.92 × \omega_{2}

\omega_{2} = 0.96 \frac{rad}{sec}

This is the value of new value of angular speed of merry go round.

6 0
3 years ago
If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______
adell [148]

Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

F = \dfrac{GMm}{r^2}............(1)

new gravitational force

F' = \dfrac{GMm}{r'^2}

now from the given condition

F' = \dfrac{GMm}{(3r)^2}

F' = \dfrac{GMm}{9r^2}

F' = \dfrac{1}{9}\dfrac{GMm}{r^2}

now, from equation (1)

F' = \dfrac{1}{9}F

\dfrac{F'}{F} = \dfrac{1}{9}

now, the change in gravitational force factor is equal to \dfrac{1}{9}

Hence, the correct answer is option A

3 0
3 years ago
How does regular exercise help older adult stay healthy?
gtnhenbr [62]
It’s helps the heart stay healthier by respiratory of the heart
5 0
3 years ago
Read 2 more answers
How would electron domains impact the magnetism of the substance?
Vanyuwa [196]
Magnetic domain structure is responsible for the magnetic behavior of ferromagnetic materials like iron, nickel, cobalt and their alloys, and ferrimagnetic materials like ferrite. ... Magnetic domains form in materials which have magnetic ordering; that is, their dipoles spontaneously align due to the exchange interaction.
4 0
2 years ago
A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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