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Arturiano [62]
3 years ago
11

A classic demonstration illustrating eddy currents is performed by dropping a permanent magnet inside a conducting cylinder. The

magnet does not go into free fall. Instead it reaches terminal velocity and can take a few seconds to drop a length of about a meter. Suppose the mass of the magnet is 70 g and width of 1.0 cm. It falls with a terminal velocity of 10 cm/s and the length of the pipe is 80 cm. The magnitude of the Joule heating from the eddy currents is approximately:________
Physics
1 answer:
kvv77 [185]3 years ago
4 0

Answer:

The correct solution is "0.69 N".

Explanation:

The given values are:

Mass of magnet,

m = 70 g

or,

   = 0.07 kg

Width,

= 1.0 cm

Velocity,

= 10 cm/s

Length of the pipe,

= 80 cm

Whenever the velocity is constant, then the net force which is acting on the magnet will be "0".

On the magnet,

The up-ward force will be:

⇒  F=mg

On substituting the values, we get

⇒      =0.07\times 9.8

⇒      =0.69 \ N

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A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that
pav-90 [236]

Answer:

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y_{b}( V₁² - V₂²)) / ( 2g( ηbμ + f_{rl} ± sin∅_{g}))

now given that the tracked is levelled, ∅_{g} = 0, also Y_{b} = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592)  = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore,  the unloaded braking efficiency is 84.6 %

7 0
3 years ago
From what expression was the word radar derived
tiny-mole [99]
RAdio Direction And Ranging
7 0
3 years ago
How much power does an Ox pulling a plow 20 m cross a field, exerting 120J of work over a period of 15 s?​
Aleksandr-060686 [28]

Answer:

8W

Explanation:

Given parameters:

Distance covered  = 20m

Work done  = 120J

Time  = 15s

Unknown:

How much power does the Ox exerts  = ?

Solution:

Power is the rate at which work is being done

  Power  = \frac{Work done }{time}  

 Power exerted by ox  = \frac{120}{15}   = 8W

3 0
3 years ago
EMERGENCY!! Doing my Physics homework now and I need help on this one please - all information provided, I'd be very VERY gratef
IrinaK [193]

(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)

(b) Vertical:

v_y = 30\sin 40^\circ = 19.3\frac{m}{s}

Horizontal:

v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}

(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):

s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s

The ball will be in the air for about 3.9 s.

(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:

s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m

The range is 89.7 meters.

6 0
3 years ago
Help pls!
Yuki888 [10]

A series circuit is a closed circuit in whicj the current flows through one path.

8 0
3 years ago
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