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Feliz [49]
3 years ago
9

What does an electromagnet have that a bar magnet does not ?

Physics
1 answer:
Anon25 [30]3 years ago
4 0
The bar magnet and the electromagnet act identical. The difference being a electromagnet is a coil of wire that has a power source connect to both ends, this energizes the coil with an electromagnetic field.
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How many times should the power develop by the engine of a ship increases to double is velocity i the resistance of the water to
Setler79 [48]

Answer:

If the ship speed is doubled, then the power developed is 8 times the initial value.

Explanation:

ship power is roughly proportional to the cube of the speed, so

P ∝ v³

If the speed is doubled, then the power developed becomes

P  ∝ (2)³ = 8 times

Therefore, if the ship speed is doubled, then the power developed is 8 times the initial value.

4 0
2 years ago
Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height require
strojnjashka [21]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

 

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

3 0
2 years ago
How can you relazie a perfect balck body in pratice​
Mazyrski [523]
A perfect black body can’t be realized
5 0
2 years ago
As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th
Fofino [41]

Answer:

5.04\cdot 10^8 A

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

\phi = 3.950\cdot 10^{-19}J

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

E=\frac{hc}{\lambda}

where

h is the Planck's constant

c is the speed of light

\lambda is the wavelength of the photon

Using E=\phi and solving for \lambda, we find the maximum wavelength of the radiation that will eject electrons from the metal:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m

And since

1 angstrom = 10^{-15}m

The wavelength in angstroms is

\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A

3 0
3 years ago
What is the acceleration of a 31 kg object pushed with a force of 42N
Ray Of Light [21]
I belive it could be 6.5 but I could be wrong
3 0
2 years ago
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