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3 years ago

I have a question regarding friction in rolling without slipping.

Physics

1 answer:

Solnce55 [7]3 years ago

4 0

Explanation:

They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.

(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)

In this case, you can't ignore friction because the disk wouldn't roll without it.

As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.

(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)

Hopefully this helps clear up the misunderstanding for you.

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What is the energy equivalent of an object with a mass of 1.83 kg?

xxMikexx [17]

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
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3 years ago

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The birth defect caused by the Zika Virus to unborn infants is called__________?

lions [1.4K]

It is B, Microcephaly. Hope I helped! :))

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3 years ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

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3 years ago

A 90 kg body is taken to a planet where the acceleration due to

Serggg [28]

Answer:

2250N

Explanation:

W= mg,

where W= weight

m= mass

g= acceleration due to gravity

Given that the body is 90kg, m= 90kg.

Acceleration due to gravity of planet

= 2.5(10)

= 25 m/s²

Weight of body on planet

= 90(25)

= 2250N

*Mass is the amount of matter an object has and is constant (same on earth and the planet).

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3 years ago

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A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

The height of water in the column = 348.14 cm

Explanation:

Pressure:This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²

p = Dgh............... Equation 1

Where p = pressure, D = density, g = acceleration due to gravity, h = height.

From the question, the same pressure will support the column of mercury and water.

p₁ = p₂

Where p₁ = pressure of mercury, p₂ = pressure of water

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

h₂ = 348.14 cm

The height of water in the column = 348.14 cm

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3 years ago