1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ket [755]
3 years ago
15

A grindstone in the shape of a solid disk with a diameter of 1.0 m and a mass of 50.0 kg, is rotating at 900 rev/min. A tool is

pressed against the rim with a normal force of 200.0 N, and the grindstone comes to rest in 10.0 s. Find the coefficient of friction between the tool and the grindstone. Neglect friction in the bearings.
Physics
1 answer:
monitta3 years ago
8 0

Answer:

oh for real?

Explanation:

The solubility of glucose at 30°C is

125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.

You might be interested in
Identify a situation in which you would want to have a high
Andreyy89

Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.

Explanation:

A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.

On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.

7 0
3 years ago
Read 2 more answers
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
The muscular system brings what to the body?
kow [346]
The muscular system brings strength and endurance to the body. It helps us perform everyday activities. As well as soaks up water to keep us hydrated longer.
6 0
3 years ago
The law of reflection states that if the angle of incidence is 32 degrees, the angle of reflection is ___ degrees.
shusha [124]

The angle of reflection is  equal to angle of incidence so the angle of reflection is also 32°.

6 0
3 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
Other questions:
  • If an asteroid is 4 AU from the Sun, what is the period of revolution around the Sun for the asteroid?
    13·1 answer
  • A bus accelerated at 1.8 m/s2 from rest for 15 s. It then traveled at constant speed for 25 s, after which it slowed to a stop w
    15·1 answer
  • "Close the door! We aren't heating the outdoors!" is a phrase my father used, especially in the winter. As a matter of fact, all
    10·1 answer
  • A bond between a positively-charged particle and a negatively-charged particle is a(n) _______ bond.
    6·1 answer
  • Physics question: Momentum
    6·1 answer
  • The number of particles in a gas system inversely affects _____ and directly affects _____.
    10·1 answer
  • jonatha want to put ketchup o​​​​​n his hamburger.he truns the ketchup bottle at an angle toward his plate and smacks the bottom
    14·1 answer
  • Which is the best term to describe a chemical reaction in which the reactants have less potential energy than the products?
    15·1 answer
  • Which of the following is TRUE regarding the development and growth of new neurons in the human brain?
    14·1 answer
  • If a star were located exactly at each celestial pole, the corrected altitude of the star would equal __________.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!