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NARA [144]
3 years ago
13

A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp

inning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.2. The stone slows steadily to 180 rpm in 10 s of grinding.
a) What is the magnitude of the angular acceleration of the grindstone as it slows down?

b) With what force does the tradesman press the knife against the grindstone?
Physics
1 answer:
Lorico [155]3 years ago
7 0

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2

b) Assume the grind stone is a solid disk, its moment of inertia is

I = mR^2/2

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

I = 28*0.15^2/2 = 0.315 kgm^2

So the friction torque is

T_f = I\alpha = 0.315*0.21 = 0.06615 Nm

The friction force is

F_f = T_f/R = 0.06615 / 0.15 = 0.441 N

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

N = F_f/\mu = 0.441/0.2 = 2.205 N

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Answer:

mechanical

Explanation:

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3 0
2 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo
Helga [31]

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

6 0
3 years ago
A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

6 0
3 years ago
The effective nuclear charge experienced by the outermost electron of Na is different than the effective nuclear charge experien
Nesterboy [21]

Answer:

B) Na has a lower first ionization energy than Ne.  

Explanation:

The atomic number¹ for Na has a value of 11 while in the case of Ne this value is 10. That means that Sodium (Na) has a total number of 11 protons, 11 neutrons and 11 electrons (since it is electrically neutral²). For the case of Neon (Ne) it has 10 protons, 10 neutrons and 10 electrons.

As the atomic number increases, the atomic radius³ shrinks (the orbitals are closer to the nucleus) as a consequence of the electric force. For the case of sodium (Na) the electron in the outermost orbital will experience a lower electric force than the electron placed in the outermost orbital in the atom of Neon (Ne).

Although, the sodium’s atom has more protons and therefore electrons, these eleven electrons will be organized according with the electronic configuration⁴ in the different shells (orbitals) of probabilities of their positions around the atom.

The electronic configuration for Na is:

1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

1s²2s²2p⁶

Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

F = \kappa_{0} \frac{q1q2}{r^{2}}    (1)

Where q1 and q2 are the charges, \kappa_{0} is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (<u>for that case will be the first ionization energy</u>).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10

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