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.
Answer:
resultant force = (f1²+f2²)½
=(1.5²+2²)½
=(2.25+4)½
=(6.25)½
=2.5
Explanation:
okay this question seems easy. now if the 1.5 is vertically upwards so is that 2 is horizontally downwards hence as u say its 90 degrees thn it forms a right angled triangle.
I think it was either Robert Hook or Anton Van Leevwen Hoek
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = 
= - 
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object
Answer:
5.77×10¹⁰ m
Explanation:
From the question,
Applying
Kepler's third law
P² = d³...................... Equation 1
Where P = Planet's period, d = distance between the center of the planet and the sun.
make d the subject of the formula n equation 1
d = ![\sqrt[3]{P^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7BP%5E%7B2%7D%20%20%7D)
.................. Equation 2
Given: P = 0.24 sidereal.
Substitute the value of P into equation 2
d = ![\sqrt[3]{0.24^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.24%5E%7B2%7D%20%7D)
d = 0.386 Au
d = 0.386×1.496×10¹¹
d = 5.77×10¹⁰ m
