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3 years ago

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21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a

speed of 660. m/s over Earth’s south magnetic pole, where the 8.00 × 10−5 − T magnetic field points straight down into the ground. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

1 answer:

beks73 [17]3 years ago

6 0

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

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You have been training and can run a mile in 6.5 minutes. How long would it take to run a lap around the outside track, which is

Paha777 [63]

Answer:

1 minute 36.85 seconds

Explanation:

First we need to convert the miles into meters, as the demanded result should be in meters.

1 mile = 1,609.34 meters

Also, 6.5 minutes should be converted into seconds.

1 minute = 60 seconds

6.5 x 60 = 390 seconds

Now we need to divide the miles with the seconds to see how much meters have been run in a second.

1,609.34 / 390 = 4.13 meters

The suggested meters now should be divided with the distance run in one second.

400 / 4.13 = 96.85 seconds

So we get a result of 96.85 seconds, or 1 minute 36.85 seconds.

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2 years ago

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

bixtya [17]

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

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2 years ago

What does a cell use to eliminate a substance that is too large to leave by diffusion?

zysi [14]

<span>a cell eliminates endocytosis.

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3 years ago

How much kinetic energy does a 1500 kg car have if it’s moving at 15 m/s

Tresset [83]

Answer:

Mass of car m = 1500 kg

velocity v = 15 m/s

kinetic energy = ½ mv2

= ½ x 1500 x (15)2

= 1687501 kg m2 /s2

= 168750J

Explanation:

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago