Answer:
yes
Explanation:
objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
Answer:
x = 2000 Km
Explanation:
Given
y = 10 km
Slope: 1 : 200
x = ?
We can apply the formula
y / x = 1 / 200 ⇒ x = 200*y = 200*10 Km
⇒ x = 2000 Km
Answer:
Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...
Explanation:
Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.
For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc
If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc
Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...
Answer:
According to the Conservation of Momentum,
Momentum of the gun = momentum of the bullet
M(gun)×V(gun)=m(bullet)×v(bullet)
4kg × V = 0.3kg × 600m/s²
V = (0.3 × 600)/4 = 45 m/s
The recoil velocity on the gun is <em><u>45 m/s</u></em>
<h3><u>45 m/s</u> is the right answer.</h3>
