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Viktor [21]
3 years ago
15

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

wheel is rotated through angles of 21 rad and 21 rev, the same point travels through arc lengths of 120 m and 7.6×102 m, respectively. What is the radius of the wheel? Answer in units of m.
Physics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

angle \times Radius = Arc

now we know that

Arc = 2.1 m

Angle = 21 degree

Angle = 21\times \frac{\pi}{180}

so now we have

21\times \frac{\pi}{180} = \frac{2.1}{R}

R = \frac{2.1 \times 180}{21 \times \pi}

R = 5.73 m

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JulsSmile [24]

Answer:

Choice a. 70\; \rm N, assuming that the skating rink is level.

Explanation:

<h3>Net force in the horizontal direction</h3>

There are two horizontal forces acting on the boy:

  • The pull of his friend, and
  • Frictions.

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N.

<h3>Net force in the vertical direction</h3>

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

<h3>Net force</h3>

Therefore, the (combined) net force on this boy would be:

\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N.

4 0
3 years ago
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50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his
DerKrebs [107]

The ball's horizontal and vertical velocities at time t are

v_x=v_{xi}

v_y=v_{yi}-gt

but the ball is thrown horizontally, so v_{yi}=0. Its horizontal and vertical positions at time t are

x=v_{xi}t

y=20.0\,\mathrm m-\dfrac g2t^2

The ball travels 22 m horizontally from where it was thrown, so

22\,\mathrm m=v_{xi}t

from which we find the time it takes for the ball to land on the ground is

t=\dfrac{22\,\rm m}{v_{xi}}

When it lands, y=0 and

0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2

\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}

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3 years ago
Please help what is the definition ligthspeed
aev [14]

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The speed of light is constant and is exactly: 299,792,458 meters per second

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3 years ago
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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr
Illusion [34]

Answer:

F_{sprinter}=110.4N

Explanation:

Given data

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Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

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From work-energy theorem we know that the work done equals the change in kinetic energy

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And

W=F_{total}.d

So

W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

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