In order to determine the acceleration of the block, use the following formula:
Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:
Then, you have:
by solving for a, you obtain:
In this case, you have:
k: spring constant = 100N/m
m: mass of the block = 200g = 0.2kg
x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m
Replace the previous values of the parameters into the expression for a:
Hence, the acceleration of the block is 10 m/s^2
Answer:
1.52g
Explanation:
Given parameters:
Force = 125N
Mass combined = 82kg
Unknown:
Acceleration of the bicycle = ?
Solution:
From Newton second law of motion suggests that:
Force = mass x acceleration
Acceleration = 
= 
= 1.52g
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m
Substitute these values into...
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N
Therefore, your answer should be <span>2.6 × 10–10</span>
Edit: You do mean Ridge?
Rocks near Mid-Ocean Ridge are younger than rocks near the trenches.
Seismic data shows oceanic crust is sinking into the mantle at the trenches.
Matching bands of magnetic rock are found on either side of the Ridge. Earth's magnetic fields change these bands over time.
