Answer:
When extra energy is added
Explanation:
When the ball is released from rest and swings back towards your face, it will only pass closer to the end of the nose as per the initial conditions. However, when extra energy is added to the ball, it strikes the nose since its velocity and heights are increased. Therefore, the only condition under which the ball hits your nose is when extra energy is added to the system.
Answer:
12900 W
24200 W
Explanation:
Given:
v₀ = 0 m/s
v = 1.3 m/s
t = 2.0 s
Find: a and Δx
v = at + v₀
(1.3 m/s) = a (2.0 s) + (0 m/s)
a = 0.65 m/s²
Δx = ½ (v + v₀) t
Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)
Δx = 1.3 m
While accelerating:
Newton's second law:
∑F = ma
F − mg = ma
F = m (g + a)
F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)
F = 19855 N
Power = work / time
P = W / t
P = Fd / t
P = (19855 N) (1.3 m) / (2.0 s)
P ≈ 12900 W
At constant speed:
Newton's second law:
∑F = ma
F − mg = 0
F = mg
F = (1500 kg + 400 kg) (9.8 m/s²)
F = 18620 N
Power = work / time
P = W / t
P = Fd / t
P = Fv
P = (18620 N) (1.3 m/s)
P ≈ 24200 W
Explanation:
The SI unit of measurement for speed is metres per second (m/s) .
Speed : The distance travelled by a body per unit time is know as speed.
Hope it will help.
Answer:
it lets out the air inside the balloon and I´m pretty sure it would hit the wall while the air was being let out.
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.
