Answer:
Explanation:
When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .
Hence normal force = reaction force .
From second law also net force is zero , so if normal force is N and reaction force is R
R - N = mass x acceleration = mass x 0 = 0
R = N .
Ranking normal force from highest to smallest
150 N , 130 N , 120 N
B )
Frictional force is equal to the weight of the body because the body is held at rest .
Ranking of frictional force form largest to smallest
7 kg , 5 kg , 3 kg , 1 kg .
Here frictional force is irrespective of the normal force acting on the body because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .
Answer:
From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to
1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.
2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity
3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.
4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.
5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.
G = 9.81 m/sec^2) g = 9.81
<span>Solving for velocity : </span>
<span> = 2gh </span>
<span>v = </span>
<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
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.
