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
larisa86 [58]
3 years ago
12

A turntable has an angular velocity of 3.5 rad/s. A dust bunny is on the disk of the turntable at a distance of 0.2m from the ce

nter of rotation. The coefficient of static friction required to keep the dust bunny from getting slung off is at leastA. 0.51B. 0.45C. 0.33D. 0.12E. 0.25
Physics
1 answer:
diamong [38]3 years ago
3 0

Answer:

E. 0.25

Explanation:

Given that

Angular speed ω = 3.5 rad/s

Distance ,r= 0.2 m

Lets take mass of dust bunny = m

We know that

Radial force F = m ω² r

The friction force on the dust bunny Fr

Fr= μ m g

To getting slung off  dust bunny from disk

m ω² r =  μ m g

ω² r =  μ  g

3.5²  x 0.2 =  μ x 10                        ( take g =10 m/s²)

μ = 0.245

μ = 0.25

Therefore answer is E

E. 0.25

You might be interested in
A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center?
anygoal [31]

Answer:

A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center? [0.297 m/s] 6. A merry-go-round a.k.a “the spinny thing” is rotating at 15 RPM, and has a radius of 1.75 m A.

5 0
3 years ago
Read 2 more answers
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step
satela [25.4K]

Answer:

W = 16.5 Kj

P = 49.9 Watt

E = 16471

Explanation:

m = 73.5kg

t = 5min 30sec = (5×60) + 30 = 330sec

each step = 16.6cm = 0.166m

h = 135×0.166 = 22.41 m

g = 10 m/s²

(i) W = F × s = W × h = mgh

W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0
3 years ago
An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
4 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

X \rightarrow X' + \alpha

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

n \rightarrow p + e^- + \bar{\nu}

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

A = 211

So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

p \rightarrow n + e^+ +\nu

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

X' \rightarrow X + \gamma

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

8 0
4 years ago
A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls without slipping down an inclined plane of height 5.3 m. what is t
inn [45]
<span>Answer: Total kinetic energy at the bottom = 0.5(1+0.4) mv^2 = mgh V^2 = 7*9.8/0.7 V = 9.9m/s ω = V/r = 9.9/1.7 = 5.8rad/s Answer c. 5.8 rad/s</span>
7 0
4 years ago
Other questions:
  • _______ currents assist in the transfer of thermal energy from warmer climates to cooler climates.
    6·1 answer
  • Why is the answer D?
    10·1 answer
  • An object has a weight of 950 Newtons, what is its mass? Round to the nearest tenth (one place after the decimal).
    15·1 answer
  • How could you dissolve more solid solute in a saturated solution in a liquid solvent
    6·1 answer
  • During a physical science lab investigating chemical reactions, several students placed a 30g antacid tablet in a 30g zip-lock b
    7·2 answers
  • ASAP
    10·1 answer
  • What about the atomic model shown below was shown incorrect?
    6·1 answer
  • The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part
    10·1 answer
  • The diagram shows a person holding a bow and arrow.
    10·2 answers
  • What type of device forms images by changing the speed at which light
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!