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A boy exerts an unknown horizontal force as he pulls a 52 N sled across packed snow. The coefficient of friction is 0.12. If a p

Tems11 [23]

Answer:

is there a pic?

Explanation:

7 0

2 years ago

Read 2 more answers

Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.

skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

= 0.5 * 1000 * (20)^2

= 0.5 * 1000 * 400

= 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

6 0

2 years ago

A negatively charged particle is attracted to

Tatiana [17]

Answer: I think the answer is D.

Positively charged particles.

8 0

3 years ago

A small loop of area A = 5.5 mm2 is inside a long solenoid that has n = 919 turns/cm and carries a sinusoidally varying current

Tema [17]

Answer: $410.90\times 10^{-6} V$

Explanation:

Given

$A=5.5 mm^2$

$n=919 turns/cm\approx 85400 turns/m$

$\omega =226 rad/s$

According to the Faraday law of induction, induced emf is given by

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

Magnetic field $\phi _B=B\cdot A$

$B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )$

$B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)$

$B=0.3305\sin (226t)$

$\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}$

$\phi _{B}=1.818\times 10^{-6}\sin (226t)$

$E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$E=-1.818\times 10^{-6}\times 226\cos (226t)$

$E=-410.90\times 10^{-6}\cos (226t)$

Amplitude of EMF induced $=410.90\times 10^{-6} V$

8 0

4 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

7 0

3 years ago

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