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3 years ago

Referring to the diagram above, predict what will happen when the switch is closed. Explain your answer.

Physics

1 answer:

alisha [4.7K]3 years ago

7 0

Answer

The nail will become an electromagnet.

Explanation

When the switched is closed , a current will flow in the insulated wire. Then the current flows through the coil around the iron nail, it will induce am magnetic filed inside the nail. The nail becomes a magnet because it is made of magnetic material. As long as there is current flowing through the current, there nail will remain a magnet.

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4) The components of Vector A are given as follows Ax = -2.4 Ay = + 3.8 What is the magnitude of the A, and what is the angle th

Monica [59]

Answer:

A = 4.49

α = 57.72°

Explanation:

Knowing the magnitude of x & y of a vector we can determine the total magnitude of a vector.

$A=\sqrt{A_{x}^{2} +A_{y}^{2} } \\A=\sqrt{(2.4)^{2} +(3.8)^{2} }\\A=4.49$

The angle tangent can be used to determine the angle.

$tan(\alpha )=\frac{3.8}{2.4}\\tan(\alpha ) =1.5833\\\alpha =tan^{-1}(1.5833) \\\alpha = 57.72 (deg)$

5 0

3 years ago

If the height of the ramp is increased then the marble's speed will be greater because gravity will have less resistance when pu

Alik [6]

If you increase the steepness of the ramp, then you will increase the acceleration of a ball which rolls down the ramp. This can be seen in two different ways:

1) Components of forces. Forces are vectors and have a direction and a magnitude. The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the gravitational force which points along the direction of the ball's motion can accelerate the ball. The other component pushes the ball into the ramp, and the ramp pushes back, so there is no acceleration of the ball into the ramp. If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp. If the ramp is vertical, the ball just drops with acceleration due to gravity. These arguments are changed a bit by the fact that the ball is rolling and not sliding, but that only affects the magnitude of the acceleration but not the fact that it increases with ramp steepness. 

2) Work and energy. The change in potential energy of the ball is its mass times the change in height (only the vertical component counts -- horizontal displacements do not change gravitational potential energy) times the local gravitational acceleration g. This loss of gravitational potential energy shows up as an increase in kinetic energy. If the ball falls a farther distance vertically, it will have a greater kinetic energy and be going faster. Again, the kinetic energy is shared between the motion of the ball going somewhere, and the rotation of the ball, and so the details of the acceleration depend on the ball (is it hollow or solid?), but the dependence on the steepness of the ramp is the same.

7 0

3 years ago

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but

Usimov [2.4K]

Answer:

a) 4.33 pC b) 5.44*10² N/C

Explanation:

a) The vertical deflecting plates of an oscilloscope form a parallel-plate capacitor.

The value of the capacitance, for a parallel-plate capacitor with air dielectric, can be found to be as follows, applying Gauss' law to the surface of one of the plates, and assuming a uniform surface charge density:

C = ε₀*A / d

where ε₀ = 8.85*10⁻¹² F/m, A = (0.03m)², and d = 0.046 m (we assume that the informed value of 4.6 m is a typo, as no oscilloscope exists with this separation between plates).

Replacing by these values, we find the equivalent capacitance of the plates, as follows:

$C = \frac{8.85e-12F/m*(0.03m)^{2} }{0.046m} =1.73e-13 F = 0.173 pF$

By definition, the capacitance of any capacitor can be expressed as follows:

$C =\frac{Q}{V}$

where Q= charge on any of the plates, and V= potential difference between them.

As we know C and V, we can find Q as follows:

Q = C*V = 0.173*10⁻¹² F * 25.0 V = 4.33*10⁻¹² C = 4.33 pC

b) We can find the electric field in several ways, but one very easy is applying Gauss' Law to a pillbox with a face outside one of the plates (paralllel to it) and the other inside the surface.

The total electric flux through the surface must be equal to the enclosed charge, divided by ε₀.

If we look to the flux crossin any face, we find that the only one that has a non-zero flux, is the one outside the surface.

As the electric crossing the boundary must be normal to the surface (in electrostatic conditions, no tangential field can exist on the surface) , and we assume that the surface charge density that creates it is constant across the surface, we can write the Gauss ' Law as follows:

E*A = Q / ε₀

where A = area of the plate = (.03m)² = 9*10⁻⁴ m², Q= charge on one of the plates = 4.33*10⁻¹² C (as we found in a)) and ε₀ = 8.85*10⁻¹² N/C.

Replacing by these values, and solving for E, we have:

$E = \frac{4.33e-12C}{(0.03m)^{2} 8.85e-12F/m} =5.44e2 N/C$