The correct answer is:
<span>B.) At terminal velocity there is no net force
In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>
<span>because the acceleration is zero: a=0.
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
Answer:
1/8
Explanation:
17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...
(1/2)(1/2)(1/2) = 1/8
1/8 of the sample remains after 17,100 years.
Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
, 
- Initial and final speeds of the shell, measured in meters per second.
If we know that 
, 
, 
and 
, then the explosive force experienced by the shell inside the barrel is:
![F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%281250%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%28750%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D%7D%7B2%5Ccdot%20%2815%5C%2Cm%29%7D)
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
Answer:
The answer of the part (a) is v2 = 7.09 m/s
and the answer of the part (b) is vA1 = 5.25 m/s
Explanation:
Explanation of the both parts of answer is in the following attachments
