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dexar [7]
2 years ago
7

An electromagnet on the ceiling of an airplane holds a steel ball. When a button is pushed, the magnet releases the ball. The ex

periment is first done while the plane is parked on the ground, and the point where the ball hits the floor is marked with an X. Then the experiment is repeated while the plane is flying level at a steady 500 mph.Does the ball land slight in front of the X( toward the nose of the plane), on the X, or slightly behind the X(toward the tail) explain.
Physics
1 answer:
blagie [28]2 years ago
7 0

Answer:

The ball will fall on the X .

Explanation:

At height,  when the aeroplane is in great speed , everything attached with it acquires the same speed . So ball will also have the same speed as the aeroplane have. When ball  starts falling off , it gets detached from plane but , at the same time it continues to travel with its earlier speed , because of inertia of motion. So it remains stationary with respect to plane in horizontal direction . It has velocity with respect to plane only in vertical direction. Hence it will fall on the X. It is due to first law of motion.

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Answer: true

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
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Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
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3 years ago
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109
Oksana_A [137]

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 A) F = 1.09 10 5 N, b) Yes  

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper +1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

               ρ = m / V

               .m = ρ V = ρ 4/3 π r³

The radius is half the diameter

               r = 1.9 10⁻² / 2 = 0.95 10⁻² m

               ρ = 8960 kg / m3

               m = 8960 4/3 π (0.95 10⁻²)³

               m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

             #_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

             #_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

               q = e / 10⁹    #_atom

               q = e / 10⁹    3,049 1023

               q = 3,049 10⁴  (-1.6 10⁻¹⁹)

               q = -4,878 10-5 C

Electric force is

             F = k q₁q₂ / r²

             F = k q² / r²

             

Let's calculate

             F = 8.99 10⁹ (4.878 10⁻⁵)²2 / (1.4 10⁻²)²

              F = 1.09 10 5 N

This is a force of repulsion.

Part B

 The magnitude of this force is  in very easy to detect

4 0
3 years ago
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