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1 month ago

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Planet X has a magnetic field with strength 0.80 G in the southern direction. When a probe is placed 30 mm east of a vertical wi

re it measures a field of 0.60 G in the southern direction. What would a probe read if placed 10 mm east of the vertical wire?
A. 0.20 G towards the south
B. 0.60 G towards the north
C. 1.00 G towards the south
D. 1.40 G towards the north

1 answer:

kicyunya [3.2K]1 month ago

8 0

0.20 G directed southward

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

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Answer:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$ ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + $z^{2}$ k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu$

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + $v^{2}$ k

The normal vector N is computed as

$N = S_{u}XS_{v}$

Where:

$S_{u} = =$

$S_{v} = =$

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, $v^{2}$ >. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + u $v^{2}$

Thus

$\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu$ ≈ 3077.34

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24 days ago

A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa

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A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.

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1 month ago

A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o

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Response:

The temperature is $T = \frac{h}{H_O \alpha_{steel} } + T_O$

Justification:

According to the information provided in the question,

The height on a cool spring day measures $H_O$

The temperature for a cool spring day registers as $T_O$

The height variation between a cool spring day and a summer day is accounted as h

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The equation outlining the linear expansion for the steel building is given as

$h = H_o \alpha_{steel} [T-T_O]$

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13 days ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

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Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

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Using formula of discharge

$Q=\dfrac{V}{t}$

Insert the values into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We will compute the coefficient of permeability

Applying the formula for permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability measures as $8.6\times10^{-3}\ cm/s$ .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

$v=ki$

$v=\dfrac{kh}{l}$

Substituting the values into the equation

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

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1 month ago

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kicyunya [3294]

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

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Using the conservation of energy principle, we have

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which yields v = 6.8 m/s

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1 month ago