Answer:
1 joule = 0.737 foot-pound
Joule is the unit of work.
1 J = 1 N·m
In SI units
1 J = 1 kg· m/s²
0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.
Answer:
spring deflection is x = (v2 / R + g) m / 4
Explanation:
We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration
Let's write the equations on the Y axis of this description
Fe - W = m 
Where Fe is elastic force, W the weight and
the centripetal acceleration. The elastic force equation is
Fe = - k x
4 (k x) - mg = m v² / R
The four is because there are four springs, R is theradio of dip
We can calculate the deflection (x) of the springs
x = (m v2 / R + mg) / 4
x = (v2 / R + g) m / 4
Answer:
F = 352 N
Explanation:
we know that:
F*t = ΔP
so:
F*t = M
-M
where F is the force excerted by the wall, t is the time, M the mass of the ball,
the final velocity of the ball and
the initial velocity.
Replacing values, we get:
F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)
solving for F:
F = 352 N
Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U