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laiz [17]
3 years ago
9

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min. After the power is shut off, a woman continu

es to sharpen her ax by holding it against the grindstone for 9 s until the grindstone stops rotating.
(a)What is the angular acceleration of the grindstone?
(b) What is the torque exerted by the ax on the grindstone? (Assume constant angular acceleration and a lack of other frictional torques.)
Physics
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

0.186 N-m        

Explanation:

mass of the grindstone, m=1.7 kg

radius, r=8 cm

Frequency, f=730 rev/min = 12.16 rev/s

time, t=9s

final angular velocity, \omega=0

Initial angular velocity,

\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s

Angular acceleration of the grind stone is:

\alpha=\frac{\omega-\omega_o}{t}\\\Rightarrow \alpha =\frac{0-76.36}{9} = -8.48 rad/s^2

Moment of inertia:

I=mr^2+mr^2=2mr^2

I=2\times 1.7 kg\times (0.08m)^2= 0.022kg-m^2

Torque exerted by the ax on the grind stone is:

\tau=I\alpha\\\tau=0.022\times (-8.48) \\\tau=0.186N-m

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A motorcycle begins at rest and accelerates uniformly S7.9 we want to find a time to take the motorcycle to reach a speed of 100
Len [333]

The motorbike reaches 100 km/h in 3.5 seconds

Explanation:

The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the motorbike in this problem,

u = 0 (it starts from rest)

v = 100 km/h = 27.8 m/s is the final velocity

a=7.9 m/s^2 is the acceleration

Solving for t, we find the time it takes for the bike to reach that velocity:

t=\frac{v-u}{a}=\frac{27.8-0}{7.9}=3.5 s

Learn more about accelerated motion:

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6 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
3 years ago
Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance
jeka94
Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N
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3 years ago
Can yall answer this? (for class) whats the diffrentce between drug addict and acoholic?
lora16 [44]

Answer:none

Explanation:

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