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3 years ago

Dale skis down a hill with a slope of 30°. Given that there is friction acting

Physics

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

The answer is A.

Explanation:

zhenek [66]3 years ago

7 0

Answer

For Apex, the answer is B.

Explanation

The arrow labeled "weight" always points down in these diagrams.

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What is the name of Newton's first law of motion? A. law of inertia B. law of acceleration C. law of objects at rest D. law of o

givi [52]

The answer is a.......

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3 years ago

Read 2 more answers

A simple dipole consists of two charges with the same magnitude, q, but opposite sign separated by a distance d. The EDM (electr

DedPeter [7]

Answer:

a. dW = ∫pEsinθdθ b. W = p.E

Explanation:

a. We know torque τ = p × E = pEsinθ where θ is the angle between p and E

Let the torque τ rotate the dipole by an amount dθ. So, the workdone dW = ∫τdθ = ∫pEsinθdθ

b. So, the total work done is gotten by integrating from 90 to θ. So,

W = ∫₉₀⁰dW

= ∫₉₀⁰pEsinθdθ

= pE∫₉₀⁰sinθdθ

= pE(cosθ - cos90)

=pEcosθ

= p.E

8 0

3 years ago

Magnet attracts nails but not copper vessels

Inessa [10]

Copper is a metal but it is not magnetic like a magnet

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2 years ago

12. The diameter of a circle is 2.42m. Calculate its<br>area in proper significant figure

Sever21 [200]

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

$A=\frac{\pi }{4} *D^{2}$

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

$A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]$

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2 years ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

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3 years ago