Answer:
Explanation:
The equation used to determine the maximum height of the bowling pin during its trajectory is given by;
H = u²/2g
where u, the initial speed/velocity, equals 10m/s
g stands for gravitational acceleration = 9.81m/s²
Substituting in the values gives us
H = 10²/2(9.81)
H = 100/19.62
Consequently, the highest point of the bowling pin's center of mass is approximately 5.0m.
mass₃<mass₁=mass₅<mass₂=mass₄
Explanation:
Data points:-
1. mass: m speed: v
2. mass: 4 m speed: v
3. mass: 2 m speed: ¼ v
4. mass: 4 m speed: v
5. mass: 4 m speed: ½ v
We know that the formula for Kinetic energy (KE) is ½ mv²
Where m represents the mass of the object
v represents the object's velocity
<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">
The KE of Body 1(mass₁) = ½*m*v² = mv²/2
KE of Body 2(mass₂) = ½*4m*v² = 2mv²
KE of Body 3(mass₃) = ½*2m*(1/4v)² = mv²/16
KE of Body 4(mass₄) = ½*4m*v ² = 2mv
²
KE of Body 5(mass₅) = ½*4m*(1/2v)² = mv²/2
</psubstituting>
Answer:
The answer to the specified question will be "
".
Explanation:
Referring to the question,
⇒ 
⇒
...(equation 1)

⇒ 
⇒
...(equation 2)
Now,
From equation 1 and equation 2, we conclude
⇒ 
By substituting the value of
, we derive
⇒ 
⇒ 
Response:
The intensity of light 18 feet underwater is about 0.02%
Clarification:
Employing Lambert's law
Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium
Then dI / I = kdt
Taking logarithms,
ln(I) = kt + ln C
I = Ce^kt
At t=0, I=I(0) implies C=I(0)
I = I(0)e^kt
At t=3 & I=0.25I(0), we find 0.25=e^3k
Solving for k gives k = ln(0.25)/3
k = -1.386/3
k = -0.4621
I = I(0)e^(-0.4621t)
I(18) = I(0)e^(-0.4621*18)
I(18) = 0.00024413I(0)
The intensity of light 18 feet underwater is about 0.2%
3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.