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abruzzese [7]
3 years ago
9

37. Which of the following is not used in cal-

Physics
1 answer:
Masja [62]3 years ago
6 0

Answer:

B

Explanation:

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A 2-kg bowling ball rolls at a speed of 10m/s on the ground whats its ke
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Answer:

Explanation:

I wonder what level of physics you are taking? If this is a beginning course (which I'm going to assume) then the KE is

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A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?
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A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How
Viktor [21]

Answer:

(a)v_c=1.56\ m.s^{-1}

(b)KE=5.3539\ J

(c-i)v_h=3.12\ m.s^{-1} in straight rightward direction.

(c-ii)v_l=0\ m.s^{1}

(c-iii)v_r=2.2062\ m.s^{-1}  \theta=45^{\circ} to the bottom of horizontal right.

(d-i) v_h=1.56\ m.s^{-1} to the horizontal right.

(d-ii)v_l=1.56\ m.s^{-1}  horizontally left

(d-iii)v_r=1.56\ m.s^{-1}  moving vertically downward

Explanation:

Given:

mass of hoop, m=2.2\ kg

diameter of hoop, d=1.2\ m

angular speed of hoop, \omega=2.6\ rad.s^{-1}

So, time taken for 1 revolution(2π radians) of the hoop:

T=\frac{2\pi}{2.6}\ s

(a)

The center will move linearly in the right direction.

circumference of the hoop:

c=\pi.d

c=\pi\times 1.2

<u>Now the speed of center:</u>

v_c=\frac{c}{T}

v_c=\frac{\pi\times 1.2}{(\frac{2\pi}{2.6})}

v_c=1.56\ m.s^{-1}

(b)

Moment of inertia for ring about central axis:

I=m.r^2

where 'r' is the radius of the hoop.

I=2.2\times 0.6^2

I=0.792\ kg.m^2

∴Kinetic energy

KE=\frac{1}{2} I.\omega^2+\frac{1}{2} m.v_c^2

KE=\frac{1}{2} 0.792\times 2.6^2+\frac{1}{2} 2.2\times 1.56^2

KE=5.3539\ J

(c) (i)

The highest point on the hoop will have the maximum velocity.

Given by:

v_h=\omega \times d

v_h=2.6\times 1.2

v_h=3.12\ m.s^{-1} in straight rightward direction.

(c) (ii)

Lowest point n the hoop will seem stationary for an observer on the ground.

v_l=0\ m.s^{1}

(c) (iii)

Velocity of the right-most point of the loop.

This velocity will have 2 components horizontal right and vertical down.

v_r=\sqrt{v_c^2+v_d^2}

here: v_d is the downward component.

v_d=r.\omega

v_d=1.56\ m.s^{-1}

\therefore v_r=\sqrt{1.56^2+1.56^2}

v_r=2.2062\ m.s^{-1}

tan\theta=\frac{1.56}{1.56}

\theta=45^{\circ} to the bottom of horizontal right.

When the observer is moving in the same direction with v_c velocity:

(d) i

Then,

v_h=\omega \times r

v_h=2.6\times 0.6

v_h=1.56\ m.s^{-1} to the horizontal right.

(d) ii

The bottom point of hoop will seem to move horizontally left with velocity:

v_l=r.\omega

v_l=0.6\times 2.6

v_l=1.56\ m.s^{-1}  horizontally left

(d) iii

Contrary to the case of stationary observer, this observer will see the extreme right point of the hoop moving vertically downward with a velocity:

v_r=r.\omega

v_r=0.6\times 2.6

v_r=1.56\ m.s^{-1}

5 0
3 years ago
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