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2 years ago

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A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

is the minimum coefficient of static friction between the turntable and the coin if the coin is not to slip?

1 answer:

Crank2 years ago

3 0

Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

$N$ = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

$N = mg$

$f$ = frequency of rotation

Angular velocity of turntable is hence given as

$w = 2\pi f$

$r$ = distance from the axis of rotation

$\mu$ = minimum coefficient of static friction

static frictional force is given as

$f = \mu N\\f = \mu mg$

The static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

$m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}$

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You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the res

lord [1]

Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

$A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55$

Add them all together to get the x component of the resultant vector, V:

$V_x=55$

Do the same to find the y components of the part of this journey:

$A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12$

Add them together to get the y component of the resultant vector, V:

$V_y=88$

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

$V_{mag}=\sqrt{55^2+88^2}$ and, rounding to 2 sig dig's as needed:

$V_{mag}=$ 1.0 × 10² m; now for the direction:

$\theta=tan^{-1}(\frac{88}{55})=$ 58°

7 0

2 years ago

Which process produces rising air when mountains push the air upward?

kicyunya [14]

Answer;

A. orographic lifting

Explanation;

Orographic lifting is a process that takes place when an air mass is forced from a low elevation to a higher elevation as it moves over rising terrain.

This can be explained by; When air is blocked by mountains, it cannot go through these mountains, As it ascends or moves up the mountain, the air then cools as it rises and when it cools to its saturation point, the water vapor condenses and cloud forms.

These clouds formed are known as orographic clouds, that develop in response to the lifting forced by the topography the earth.

5 0

2 years ago

Read 2 more answers

Please answer fast in hindi

Ipatiy [6.2K]

Answer:

1. Dheere Dheere (slowly slowly)

2. Har (every)

3. Kal (tomorrow)

4. Mat (don't)

5. Andar (inside)

sorry I wasn't able to write in hindi

8 0

2 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb

marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

$n_1 =\dfrac{46 \times 1}{0.0821\times 298}$

n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

$n_2 =\dfrac{12 \times 1}{0.0821\times 298}$

n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

$P_1=\dfrac{n_1 R T}{V}$

$P_1=\dfrac{1.89\times 0.0821\times 298}{5}$

P₁ = 9.25 atm

Partial pressure of oxygen

$P_2=\dfrac{n_2 R T}{V}$

$P_2=\dfrac{0.49\times 0.0821\times 298}{5}$

P₂ = 2.39 atm

total pressure

P = P₁ + P₂

P = 9.25 + 2.39

P = 11.64 atm

8 0

2 years ago