<span>Answer:
Total kinetic energy at the bottom = 0.5(1+0.4) mv^2 = mgh
V^2 = 7*9.8/0.7
V = 9.9m/s
ω = V/r = 9.9/1.7 = 5.8rad/s
Answer c. 5.8 rad/s</span>
By the law of momentum conservation:-
=>m¹u¹ + m²u² = m1v1 + m²v² {let East is +ve}
=>u¹ + u² = v¹ + v² {as m1=m2}
=>3.5 - 2.75 = v1-1.5
<span>
=>v¹ = 2.25 m/s (East) </span>
Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here,
Here,
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
The bicyclist accelerates with magnitude <em>a</em> such that
25.0 m = 1/2 <em>a</em> (4.90 s)²
Solve for <em>a</em> :
<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²
Then her final speed is <em>v</em> such that
<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)
Solve for <em>v</em> :
<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s
Convert to mph. If you know that 1 m ≈ 3.28 ft, then
(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h