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Mama L [17]
2 years ago
6

If you applied a force of 200 Newtons to lift a box 2.20 meters above the floor, how much work would you be doing?

Physics
1 answer:
Fittoniya [83]2 years ago
7 0
Hi hi hhhhhjjjjjiiiijjjj
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Static electricity occurs when electrons build up and _________
EleoNora [17]
Static electricity<span> is a </span>buildup<span> of </span>electric<span> charges on objects. Charges </span>build up<span> when negative </span>electrons<span> are transferred from one object to another. The object that gives </span>up electrons<span> becomes positively charged, and the object that accepts the </span>electrons<span> becomes negatively charged. This can </span>happen<span> in several ways</span>
5 0
3 years ago
cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your
sp2606 [1]

We have that the time in seconds, minutes, and hours is

t=1.083*10^{-4}s

T_{min}=1.805*10^{-6}min

T_{hours}=3.0083*10^{-8}hours

From the Question we are told that

Velocity v=13.0m/s

Distance d=120 km

Generally the equation for the Time  is mathematically given as

t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s

Therefore

T_{min}=1.083*10^{-4}s/60

T_{min}=1.805*10^{-6}min

And

T_{hours}=T_{min}/60

T_{hours}=3.0083*10^{-8}hours

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

4 0
3 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0
3 years ago
How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?
harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

U = \mu B

where;

  • \mu is the dipole moment
  • B is the magnetic field

B = \frac{\mu_0 I}{2\pi r}

U = \mu\times  (\frac{\mu_0 I}{2\pi r} )

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

  • In step 1, to increase the potential energy, the iron will move towards the electromagnet.
  • In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

7 0
2 years ago
A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe
Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2

Acceleration is 51.94 ft/s²

v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s

Final velocity at this time is 257.63 ft/s

5 0
3 years ago
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