Explanation:
Data provided:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Separation distance d between the plates = 1 mm = 1×10⁻³ m
Battery voltage, or emf = 100 V
Resistance = 1025 ohm
Solution:
In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as
Applying the natural logarithm to both sides,
(1)
Next, we can determine the capacitance using the plates' area.
C = ε₀A/d
= 
= 18×10⁻¹²F
We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
Provided Information:
Length of inclined plane = 8 m
Height of inclined plane = 2 m
Weight of the ice block = 300 N
Required Information:
Force needed to push ice block = F =?
Answer:
Force needed to push the ice block = 75 N
Explanation:
The required force to push this ice block up an inclined plane is given by
F = Wsinθ
where W is the weight of the ice block and θ is the angle indicated in the attached image.
Using trigonometric ratios,
sinθ = opposite/hypotenuse
where the opposite side is the height of the inclined plane and the hypotenuse is the length of the inclined plane.
Thus, sinθ = 2/8
θ = sin⁻¹(2/8)
which leads to θ = 14.48°
Therefore, F = 300*sin(14.48)
results in F = 75 N
This indicates that a force of 75 N is necessary to push the ice block on the specified inclined plane.
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09
Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489
Thus, the acceleration is approximately 5.5 ft/s²!!
