Answer:
L = 1.11 x
m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x
m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x
m
Length = ?
So,
we know that,
U = 1/2 C Δ
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ
52000 J = (0.5) x (C) x (
)
C = 0.28 F
And we also know that,
C = 
E = 8.85 x 
K = 3.7
A = 0.20 x L
d = 2.6 x
m
Plugging in the values into the formula, we get:
0.28 = 
Solving for L, we get:
L = 1.11 x
m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
Forces affect how objects move. They may cause motion; they may also slow, stop, or change the direction of motion of an object that is already moving. Since force cause changes in the speed or direction of an object, we can say that forces cause changes in velocity. Remember that acceleration is a change in velocity.
Answer:
By a factor of 1/4.
Explanation:
The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,
in which
,
represent the change in momentum and the time taken for that change.
If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by
, the following manipulation confirms the answer to this question.
![\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5C%5C%5Csmall%20F_1%20%26%3D%5Csmall%20%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B4%5CDelta%20t%7D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7D%5Ctimes%5Cbigg%5B%5Cfrac%7B%5CDelta%20%28mV%29%7D%7B%5CDelta%20t%7D%5Cbigg%5D%5C%5C%5C%5C%26%3D%5Csmall%20%5Cfrac%7B1%7D%7B4%7DF%5Cend%7Baligned%7D)
Here
is the force that was applied to the object previously.
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Frequency and wavelength are inversely proportional.
A shorter wavelength implies a higher frequency.