Solve for acceleration:
<em>a</em> = (21.4 m/s - 33.8 m/s) / (4.7 s)
<em>a</em> ≈ -2.6 m/s²
Solve for force:
<em>F</em> = (1400 kg) <em>a</em> ≈ -3700 N
The minus sign tells you the force points in the opposite direction of the car's motion. Its magnitude is always positive, so <em>F</em> = 3700 N.
Answer:
m = 9795.9 kg
Explanation:
v = 35 m/s
KE = 6,000,000 J
Plug those values into the following equation:
6,000,000 J = (1/2)(35^2)m
---> m = 9795.9 kg
Answer:
F₂= 210 pounds
Explanation:
Conceptual analysis
Hooke's law
Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:
F= K*x Formula (1)
Where;
F is the magnitude of the force applied to the spring in Newtons (Pounds)
K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)
x the elongation of the spring (inch)
Data
The data given is incorrect because if we apply them the answer would be illogical.
The correct data are as follows:
F₁ =80 pounds
x₁= 8 inches
x₂= 21 inches
Problem development
We replace data in formula 1 to calculate K :
F₁= K*x₁
K=( F₁) / (x₁)
K=( 80) / (8) = 10 pounds/ inche
We apply The formula 1 to calculate F₂
F₂= K*x₂
F₂= (10)*(21)
F₂= 210 pounds
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π
Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
= (15π rad/s)*(1 rev/2π rad)
= 7.5 rev/s
Answer: 7.5 revolutions per second.
