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Helga [31]
3 years ago
12

While discussing the effects of loose wheel bearings: Technician A says that vibration may occur while driving. Technician B say

s that the bearing may make noise. Who is correct? a. Technician A b. Neither A nor B c. Technician B d. Both A and B
Physics
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

Option "D" is the correct answer i.e. Both Technicians A & B

Explanation:

Dear,

Loose wheel bearings can be caused on any machine.

The Following can be the reasons :

  • the machine or motor is installed improperly
  • the rotor(rotating part) of machine is not balanced
  • the cover get deteriorated with time
  • Bad manufacturer

So, when the bearings will be loose then the following can happen:

  1. It will produce vibration in machine which can be judged by a vibration pen eg. SKF vibration pen
  2. The vibration will also result in noise. You can check by
  3. The machine winding if it is motor will burn-out.
  4. Bush needs to be installed if the bearing runs loose.

Therefore,  "D" is right.

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0
3 years ago
А
timurjin [86]

Answer:

i hope it will be useful for you

Explanation:

F=5.6×10^-10N

R=93cm=0.93m

let take m1 and m2 =m²

according to newton's law of universal gravitation

F=m1m2/r²

F=m²/r²

now we have to find masses

F×r²=m²

5.6×10^10N×0.93m=m²

5.208×10^-9=m²

taking square root on b.s

√5.208×10^-9=√m²

so the two masses are m1=7.2×10^-5

and m2=7.2×10^-5

8 0
3 years ago
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an
nalin [4]

Answer:

The maximum volume is 1417.87 inch^3

Explanation:

<u>Optimization Using Derivatives</u>

We have a 24x30 inch piece of metal and we need to make a rectangular box by cutting a square from each corner of the piece and bending up the sides. The width of the piece is 24 inches and its length is 30 inches

When we cut a square of each corner of side x, the base of the box (after bending up the sides) will be (24-2x) and (30-2x), width and length respectively. The volume of the box is

V=(24-2x)(30-2x)x

Operating

V=4x^3-108x^2+720x

To find the maximum value of V, we compute the first derivative and equate it to zero

V'=12x^2-216x+720=0

Simplifying by 12

x^2-18x+60=0

Completing squares

x^2-18x+81-81+60=0

(x-9)^2=21

We have two values for x

x=9+\sqrt{21}=13.58\ inch

x=9-\sqrt{21}=4.42\ inch

The first value is not feasible because it will produce a negative width (24-2(13.58))=-6.16

We'll keep only the solution

x=4.42\ inch

The width is

w=(24-2(4.42))=15.16\ inch

The length is

l=(30-2(4.42))=21.16\ inch

And the height

x=4.42\ inch

The maximum volume is

V=(15.16)(21.16)(4.42)=1417.87\ inch^3

4 0
3 years ago
Mrs. Nogaki uses constant force to push a Costco shoping cart down the long store
Liono4ka [1.6K]
Figure B. The second option.
7 0
3 years ago
A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a
IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0
3 years ago
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