Answer:
C
Explanation:
frequency = 1 / period
so
frequency = 0.25
0.25 = 1 / period
1 / 0.25 = period
4 s = period
it is not 4 Hz because period is the time taken
so the answer is C
hope this helps ,please mark it
No they say "Watch out it's the fuzz"
Answer:
The amount of Potential Energy lost in the form of Thermal Energy is equal to 41.64 MJ.
Explanation:
mass of the cart and passengers = m = 3.5*10⁴ kg
Height = h = 126.5 m
g = 9.8 ms⁻²
v = 10 ms⁻¹
At the top position the Total energy of the ride is in the form of potential energy given as:
P.E = mgh
= 3.5*10⁴*9.8*126.5
= 43389500 J
At the bottom of the drop, all the Potential Energy will be converted into K.E as the ride develops speed given as:
K.E = 0.5*m*v²
= 0.5*3.5 x 10⁴*100
= 1750000 J
The K.E is much less than the P.E energy even though all the P.E of the ride has been converted. The loss of energy is due to the formation of thermal energy which can be calculated as:
Thermal Energy lost = Total energy in the form of P.E(top) - Total energy in the form of K.E(bottom)
= 43389500 - 1750000
= 41639500 J
= 41.64 M J approx M J = mega joule = 10⁶ J
To solve for this we need to use the formula P = W/T (Power = Work/Time).
Since we do not have time, we shall switch up the formula.
Our new formula is T = W/P (Time = Work/Power).
We have 1,800,000 J of work, and 15,000 W of power.
1,800,000/15,000 = 120.
It will take 120 (insert measure of time here).
I hope this helps!
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>
