Resultant force is 400-600=-200N.
-200=60a solve for a.
Explanation:
Earth rotates in prograde mation.As viewed from the north pole star Polaris.Earth turns counterclockwise,, the north pole is point in the northern,, Hemisphere where Earth's Axis of rotation meets it's surface
Hello there!
Essentially, a control variable is what is kept the same throughout the experiment, and it is not of primary concern in the experimental outcome. Any change in a control variable in an experiment would invalidate the correlation of dependent variables (DV) to the independent variable (IV), thus skewing the results.
Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective 
= 
⇒ = 
⇒ = 295 cm
Focal length of eyepiece 
= 2.7 cm
Magnifying power of a telescope is given by,
therefore the Magnifying power of a telescope is
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
