Answer:
Explanation:
Electric field E = 4 x 10⁷ V / m
Dielectric constant k = 24
capacitance of capacitor
C = kε₀ A / d
d = plate separation
A = plate area
C = .89 x 10⁻⁶
V / d = electric field
for minimum d , electric field will be maximum
V / d = 4 x 10⁷
1930 / d = 4 x 10⁷
d = 1930 / 4 x 10⁷
d = 482.5 x 10⁻⁷ m
= 48.25 x 10⁻⁶ m
C = kε₀ A / d
.89 x 10⁻⁶ = 24 ε₀ A / d
A = .89 x 10⁻⁶ X d / 24 ε₀
A = .89 x 10⁻⁶ X 48.25 x 10⁻⁶ / 24 x 8.85 x 10⁻¹²
= 42.9 / 212.4
= .2019 m²
Answeryes they are the same
Explanation:
0.5-0.5=10 is gotta be sorry if its wrong
Data charts would use descriptive statistics to show accurate reading measured throughout the lab. Charts and graphs can also be used to show the progress and result of the lab.
Answer:
x = 0.396 m
Explanation:
The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring
Data the putty has a mass m1 and velocity vo1, the block has a mass m2
. t's start using the moment to find the system speed.
Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash
p₀ = m1 v₀₁
Moment after shock
= (m1 + m2) 
p₀ =
m1 v₀₁ = (m1 + m2) 
= v₀₁ m1 / (m1 + m2)
= 4.4 600 / (600 + 500)
= 2.4 m / s
With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring
Before compressing the spring
Em₀ = K = ½ (m1 + m2)
²
After compressing the spring
= Ke = ½ k x²
As there is no rubbing the energy is conserved
Em₀ = 
½ (m1 + m2)
² = = ½ k x²
x =
√ (k / (m1 + m2))
x = 2.4 √ (11/3000)
x = 0.396 m