Answer:
different value to l if n=5 are 0,1,2,3,4
Answer: The density of this piece of jewelry is ![8.90g/cm^3](https://tex.z-dn.net/?f=8.90g%2Fcm%5E3)
Explanation:
To calculate the density, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Mass of piece of jewellery = 130.8 g
Density of piece of jewellery = ?
Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml =
![1cm^3=1ml](https://tex.z-dn.net/?f=1cm%5E3%3D1ml)
Putting values in above equation, we get:
![\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20piece%20of%20jewellery%7D%3D%5Cfrac%7B130.8g%7D%7B14.7cm%5E3%7D%3D8.90g%2Fcm%5E3)
Thus density of this piece of jewelry is ![8.90g/cm^3](https://tex.z-dn.net/?f=8.90g%2Fcm%5E3)
Answer:
a)
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂) ![v_{f}](https://tex.z-dn.net/?f=v_%7Bf%7D)
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ - ![K_{f}](https://tex.z-dn.net/?f=K_%7Bf%7D)
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system
Answer:
Equilibrium. • When an object is in equilibrium (either at rest or moving with constant velocity), the net force acting on it zero.
Answer: L can be expressed in terms of g and f as
L = g/(2πf)^2
Explanation: Please see the attachments below