Answer:
v = 18.5 m /s
Explanation:
The velocity of the jumper will go on increasing until his weight is more than the restoring force created in the bungle.
mg = kx
50 x 9.8 = 100 X
X = 4.9 m
Total fall = 15 + 4.9 = 19.9 m
kinetic energy gained due to fall by 19.9 m
= mgh = 50 x 9.8 x 19.9 = 9751 J
energy lost due to stored elastic energy in bungie
= 1/2 k x²
= .5 x 100 x 4.9²
= 1200.5 J
Remaining kinetic energy = 9751 - 1200.5 = 8550.5 J
1/2 m v² = 8550.5
.5 x 50 x v² = 8550.5
v² = 342
v = 18.5 m /s
Answer:
Explanation:
Given
Mass of car
Force applied
time period
Acceleration associated with car
Acceleration is the change in velocity w.r.t time
thus, the change in velocity 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>
Answer:
Explanation:
We can use the following kinematics equations to solve this problem:
.
Using the first one to solve for acceleration:
.
Now we can use the second equation to solve for the distance travelled by the airplane:
(three significant figures).