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dexar [7]
3 years ago
14

Credit-Card Magnetic Strips Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000

gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 7.0A . How close can a credit card be held to this wire without damaging its magnetic strip?
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer:

14 μm

Explanation:

The magnetic field due to a long straight wire is B = μ₀i/2πr where  μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current = 7.0 A and r = distance of credit card from magnetic field.

So r = μ₀i/2πB since B = 1000 gauss = 1000 G × 1 T/10000 G  = 0.1 T

r = 4π × 10⁻⁷ H/m × 7.0 A/(2π × 0.1 T)

r = 2 × 10⁻⁷ H/m × 7.0 A/0.1 T

r = 14 × 10⁻⁷ H/m × A/0.1 T

r = 140 × 10⁻⁷ m

r = 1.4 × 10⁻⁵ m

r = 14 × 10⁻⁶ m

r = 14 μm

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A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of
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m = mass of the car moving in horizontal circle = 1750 kg

v = Constant speed of the car moving in the horizontal circle = 15 m/s

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To move in a circle . centripetal force is required which is given as

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A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th
Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration g=-9.8 m/s^2 downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

v(t) = v_0 sin \theta +gt

where

v_0 = 30 m/s is the initial velocity of the rock

\theta=53^{\circ} is the angle

t is the time

Requiring v(t)=0, we find the time at which the heigth is maximum:

0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s

The heigth of the rock at time t is given by

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

x(t) = (v_0 cos \theta) t

where

v_0 cos \theta is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

x=(30)(cos 53^{\circ})(2.44)=44.1 m

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

x(t) = (v_0 cos \theta) t

Substituting

v_0 = 30 m/s\\\theta=53^{\circ}

and t = 7.58 s we find:

x=(30)(cos 53^{\circ})(7.58)=136.8 m

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

x(t) = (v_0 cos \theta) t

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

And substituting the different times, we find:

x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m

y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m

x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m

y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m

x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m

y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m

3 0
3 years ago
The jet stream flows from east to west across the United States.<br><br><br> True False
velikii [3]
This is false. they flow west to east
8 0
3 years ago
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