Answer: The magnetic flux is 3.4 × 10^-3 Wb
Explanation: Please see the attachments below
The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>
8.33 m/s
Explain:
If you understand what 1 km = 1000 m and 1 hr = 60*60 seconds signify, you already know what they mean. If you translate 1000 m/ 3600 s to km/hr, you'll get 5 m/ 18 s, which you can multiply by 30 to get (30 x 5)/18 = 8.33 m/s.
Explanation:
It is given that,
Distance, r = 3.5 m
Electric field due to an infinite wall of charges, E = 125 N/C
We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :
It is clear that the electric field is inversely proportional to the distance. So,
E' = 291.67 N/C
So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.
The crate is in equilibrium. Newton's second law gives
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where
• <em>n</em> = magnitude of the normal force
• <em>mg</em> = weight of the crate
• <em>p</em> = mag. of push exerted by movers
• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>
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It follows that
<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N
so that the movers perform
<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J
of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)
