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icang [17]
3 years ago
10

PLEASE HELP WILL GIVE BRANLIEST IF CORRECT

Physics
2 answers:
Lerok [7]3 years ago
4 0

Answer:

4

Explanation:

X would be armature.

Y is the brush because when z moves the circuits, the brush moves with it

the commutator helps turn the circuits, so that would be z.

coldgirl [10]3 years ago
3 0

Answer:

Explanation:

  • z: commutator
  • y: brush
  • x: armature

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I think the answer is B
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2 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

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As easy as one can imagine

8 0
3 years ago
A car has an acceleration of -5 m/s^2. Describe the car’s motion
AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

7 0
3 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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