Answer:
F = 812.25 N
Explanation:
Given (convert to SI units):
m = 57 g = 0.057 kg
= -25
= 32
t = 4 ms = 0.004 s
Find acceleration:
a = = = 14250
Find force:
F = ma = 0.057(14250) = 812.25 N
I’m not quite sure what your asking, but if you need an example of an increase in potential energy, a good example would be when a ball begins to slow down after rolling down a hill. If there is another part of this question, that would be great so I could help a little more but for now I hope this helped! :-)
Answer: 0.424 s
Explanation:
The speed of the car is given by:
where
v = 5.9 m/s is the speed of the car
S = 2.5 m is the distance covered by the car
t is the time taken
Substituting the numbers and re-arranging the equation, we can find the time taken:
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>
Since direction is part of velocity, and the object is moving in a circle, its velocity is constantly changing.