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jarptica [38.1K]
3 years ago
6

What is the angular displacement of the second hand on a clock after 59 seconds?

Physics
2 answers:
tigry1 [53]3 years ago
7 0

The angular displacement of the second hand on a clock after 59 second is -6.17 rad.

Answer: Option A

<u>Explanation: </u>

As we know angular displacement is defined as the change in the displacement in a circular motion for a given angle. So, angular displacement is derived using the formula

       \text { Angular Displacement }=\frac{\text { Total displacement }}{\text { Radius of the circular path }}

Also, in a clock the second hand travel a distance of 2π in 1 minute. If for 60 s, the angular displacement is 2π. Then for 1 s, the angular displacement will be

           \text { Angular Displacement for } 1 s=\frac{2 \pi}{60}

And for 59 seconds, the angular displacement will be

          \text { Angular Displacement for } 59 s=\frac{2 \pi}{60} \times 59 \text { rad }

As we do not know the displacement of seconds hand after 59 seconds but the angular displacement will be having a negative sign as the clock hands move in clockwise direction.  So ,

         \text { Angular Displacement for } 59 s=-\frac{2 \times 3.14 \times 59}{60}=-6.17 \mathrm{rad}

DerKrebs [107]3 years ago
6 0

Answer:

a. -6.17 rad

Explanation:

60 seconds is 2π radians.  Writing a proportion:

2π / 60 = x / 59

x = 6.17

The displacement is negative because the second hand moves clockwise.

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Therefore,

\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}

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1 year ago
As you sit in a chair, in what direction (or directions) do you exert an action force?
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In short, Your Answer would be Option A

Hope this helps!
5 0
3 years ago
Read 2 more answers
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
Brianna pushes a 20 kg box with a force of 50N to achieve an acceleration of 2.5 m/s/s. In order to push a 30 kg box at the same
Elanso [62]

Answer:

She applies 75N of force

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F=ma

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At what speed, as a fraction of c , is a particle's total energy twice its rest energy
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E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
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where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
which means:
\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
\frac{v}{c}=  \sqrt{1- (\frac{1}{2})^2 }  =0.87
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
3 0
3 years ago
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