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Leto [7]
3 years ago
9

Write any two uses of magnetic force​

Physics
1 answer:
OLga [1]3 years ago
7 0

Explanation:

Any two uses of magnetic force are:-

  • It is used in factories for lifting heavy metals.
  • It is used by opthalmologist for taking out the small metallic particles from eyes .
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What is a sloping surface, like a ramp, that reduces the amount of force required to do work.
puteri [66]
If this is about simple machines the answer is: Inclined plane
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3 years ago
Which of the following is not a nutrient? Minerals, probiotics, lipids or proteins. Why?
Tju [1.3M]
Sorta stuck between minerals and lips
Minerals; Minerals are, unlike vitamins, inorganic substances. All vitamins are identified as key nutrients: the body cannot produce them and so they must be obtained through diet; minerals are similarly identified as key nutrients. Vitamins and minerals are equally essential for the healthy functioning of the body.
But then there’s lipids
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Hope I helped a little
4 0
4 years ago
Which statement below best reflects the energy of the rock shown in the diagram?
Natali [406]

Answer:

32

Explanation:

32

5 0
3 years ago
A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th
mr Goodwill [35]

elasticity stretches and can also return to it's normal size ..

8 0
3 years ago
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

Now the force is

F=kx\\ or

x=\dfrac{F}{k}\\

So here F=380 N, k=630.92 N/m

x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

So the distance is 0.602 m

6 0
3 years ago
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