Imagine a right triangle where the legs represent the horizontal and vertical lengths of the string and the hypotenuse represents the length of the string.
Let us assign some values:
x = horizontal length in feet
50 = vertical length in feet
L = length of the string in feet
Because we are modeling these quantities with a right triangle, we can use the Pythagorean theorem to relate them with the following equation:
L² = x² + 50²
We want to find an equation for the change of L over time, so first differentiate both sides with respect to time t then solve for dL/dt:
2L(dL/dt) = 2x(dx/dt)
dL/dt = (x/L)(dx/dt)
First let's solve for x at the moment in time described in the problem using the Pythagorean theorem:
L² = x² + 50²
Given values:
L = 100ft
Plug in and solve for x:
100² = x² + 50²
x = 86.6ft
Now let's find dL/dt. Given values:
x = 86.6ft, L = 100ft, dx/dt = 4ft/sec
Plug in and solve for dL/dt:
dL/dt = (86.6/100)(4)
dL/dt = 3.46ft/sec
Answer:
The wood block reaches a height of 4.249 meters above its starting point.
Explanation:
The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.
(1)
Where:
- Maximum height of the wood block, in meters.
- Initial speed of the block, in meters per second.
- Kinetic coefficient of friction, no unit.
- Gravitational acceleration, in meters per square second.
- Mass, in kilograms.
- Distance travelled by the wood block along the wooden ramp, in meters.
- Inclination of the wooden ramp, in sexagesimal degrees.
If we know that 
, 
and 
, then the height reached by the block above its starting point is:
The wood block reaches a height of 4.249 meters above its starting point.
Answer:
a) v = 13.8 m / s
, b) a = 95.49 m / s²
, c) a force that goes to the center of the carnival ride and d) μ = 0.10
Explanation:
For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics
a) reduce the magnitudes to the SI system
w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s
The equation that relates linear and angular velocity is
v = w r
v = 6.91 2
v = 13.8 m / s
b) centripetal acceleration is given by
a = v² / r = w² r
a = 6.91² 2
a = 95.49 m / s²
c) this acceleration is produced by a force that goes to the center of the carnival ride
d) Here we use Newton's second law
fr -W = 0
fr = W
μ N = mg
Radial shaft
N = m a
N = m w² r
μ m w² r = m g
μ = g / w² r
μ = 9.8 / 6.91² 2
μ = 0.10
The thermal energy of an object is defined as the total translational kinetic energy of the particles that make it up.
The answer is B.
