To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground
F = ma
We have mass = 0.2kg
and acceleration = 20 m/s^2
So..
F = (0.2)(20)
F = 4 N
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
Answer:
15.75 m/s
Explanation:
v = Velocity of the combined mass of astronaut and tools = 1.8 m/s
= Mass of astronaut = 124 kg
= Mass of tools = 16 kg
= Velocity of astronaut = 0
= Velocity of tools
As linear momentum is conserved
The velocity of the tools is 15.75 m/s
Answer:
∴ fractional compression = 1.34 × 10⁻²
Explanation:
given,
depth of Indian ocean = 3000 m
Bulk modulus of the water = 2.2 x 10⁹ N/m²
We know,
P = P₀ + ρgh
P₀ is the atmospheric pressure
P₀ = 10⁵ N/m²
ρ is the density of the water, 1000 Kg/m³
P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²
using formula,
B = P/{-∆V/V}
B is bulk modulus and { -∆V/V} is the fractional compression
∴ fractional compression = 1.34 × 10⁻²
