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iren2701 [21]
4 years ago
15

A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

ction with a frequency of 3.05 Hz. What is the maximum amplitude that the piston can oscillate without the bolt losing contact with the piston's surface?
Physics
1 answer:
densk [106]4 years ago
5 0

Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

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hichkok12 [17]
A magnetic field is actually generated by a moving current (or moving electric charge specifically). The magnetic field generated by a moving current can be found by using the right hand rule, point your right thumb in the direction of current flow, then the wrap of your fingers will tell you what direction the magnetic field is. In the case of current traveling up a wire, the magnetic field generated will encircle the wire. Similarly electromagnets work by having a wire coil, and causing current to spin in a circle, generating a magnetic field perpendicular to the current flow (again right hand rule).

So if you were to take a permenant magnet and cut a hole in it then string a straight wire through it... my guess is nothing too interesting would happen. The two different magnetic fields might ineteract in a peculiar way, but nothing too fascinating, perhaps if you give me more context as to what you might think would happen or what made you come up with this question I could help.

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3 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
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Ostrovityanka [42]
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How will the magnetic field of a straight wire carrying a constant current interact with the needle of a compass? Does the inter
Yuki888 [10]

Answer:

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Explanation:

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Right hand rule tells the direction. Place the current carrying wire in your right hand with thumb pointing the direction of current. Curl of the fingers tell the direction of current.

When the needle gets in the vicinity of the field, its poles aligns itself with the field. (previous position of the compass needle has no effect on its position in the field). The north pole and south pole will be set in the direction of magnetic field.

The distance between the needle and wire does effect the strength (accuracy) of the needle position. Strong field will create strong deflection of the needle whereas when the distance from wire increases, field weakens, thus the deflection of needle will be weak.

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Answer:

A.

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