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laila [671]
3 years ago
4

An electromagnet is created using a battery, an insulated copper wire, and an iron nail. The wire is wrapped around the nail 20

times and attached to the battery. Which pair of changes will definitely increase the strength of the electromagnet?
Physics
2 answers:
Veseljchak [2.6K]3 years ago
4 0
Think of each turn as a tiny electromagnet, and they all add up. So increasing the number of turns will increase the magnet strength. And if the current increases then the magnetic field strength increases. Current = Voltage/Resistance. So increasing the voltage will increase the current.
so if there is "Increase the voltage of the battery and increases the number of turns of the coil" then that option should be the answer 
Roman55 [17]3 years ago
4 0

^The answer is C if doing USA test prep !!!


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Well gravity is 9.8m/s so i would assume 19.6m/2 is the velocity
8 0
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. Two astronauts are 1.90 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via elec
melomori [17]

Answer:

d= 1650 km.

Explanation:

  • Assuming that sound waves travel along a straight line at a constant speed of 346 m/s, we can find the time needed for the sound to travel the distance of 1.9 m between the astronauts, just applying the definition of average velocity, as follows:

       t = \frac{\Delta x}{v} =\frac{1.9m}{346m/s} = 5.5 msec  (1)

  • The electromagnetic waves travel in free space at the same speed of light in vacuum ( since light is a electromagnetic wave indeed), i.e., 3*10⁸ m/s.
  • Applying the same formula than in (1) we can solve for the distance d, as follows:

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3 0
3 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

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