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makkiz [27]
3 years ago
7

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

es to sharpen his axe by holding it against the grindstone until it stops 10 s later What is the average torque exertes by the axe on the grindstone?
Physics
1 answer:
kolbaska11 [484]3 years ago
8 0

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

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Electrical to kinetic

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A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
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1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

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3 years ago
Which statement best explains the relationship between the electric force between two charged objects and the distance between t
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A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
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Answer: Last option

2.27 m/s2

Explanation:

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in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

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a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object
tatiyna

Answer:

D_T=18.567m

Explanation:

From the question we are told that:

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Displacement d=1.05 m

Initial time t_1=6.0s

Final Time t_2=2.5s

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Generally the equation for Distance traveled before stop is mathematically given by

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Generally the equation for Distance to stop is mathematically given by

Since For this Final section

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Therefore

Using Newton's Law of Motion

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Generally the Total Distance Traveled is mathematically given by

 D_T=d_1+d_2+d_3

 D_T=5.125m+12.392+1.05 m

 D_T=18.567m

6 0
3 years ago
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