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meriva
3 years ago
12

What does an electromagnet do?

Physics
1 answer:
butalik [34]3 years ago
8 0
An electromagnet is a device that sends electricity through a coil of wire to produce a magnetic field. This leads to a magnet that can be controlled - turned on and off with the flip of a switch, or increased or decreased in strength. The coils are often wrapped around a regular magnet to make it stronger.
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Suppose that an electromagnetic wave which is linearly polarized along the x−axis is propagating in vacuum along the z−axis. The
goldfiish [28.3K]
I think the answer to your question is B
7 0
3 years ago
An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat
Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, v_{max} = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max}  \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max}  \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s

Therefore, the maximum speed is 21.39 m/s.

3 0
3 years ago
A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv
iren2701 [21]

Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively.  Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin.  The sphere is replaced with a hoop of the same mass and radius.  Will the value of Hmin increase, decrease, or stay the same?  Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0.  Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5.  For a hoop, k = 1.  As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above.  The track makes an angle of θ above the horizontal at point C.  Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate.  Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ.  At the maximum height, the vertical velocity is 0.  During this time, the sphere is in free fall.  The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0
2 years ago
What is motion and how do we determine it
spin [16.1K]

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8 0
3 years ago
Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is
Andreas93 [3]

Answer:

The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.

Examples of Gravitational Energy

A raised weight.

Water that is behind a dam.

A car that is parked at the top of a hill.

A yoyo before it is released.

5 0
2 years ago
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