Yes, scientific method can be applied on many everyday activities to get a reasonable solution. Infact normally we are applying this method without having it in our knowledge that we are applying it.
For example: In morning we are going to office and we start the car, but it is not started.You turn the engine again and again but it simply donot works.
Observation (the state of defining a problem):
The car is not started
Hypothesis (A possible solution based on the information we already know):
The car is not started because it might be out of gas or there can be some other technical fault.
Experiment (testing of hypothesis by applying different methods of solving problem):
You get the fuel and put it inside the car but it still donot works and car didnot start. Experiment didnot get solution.
Analyze the results of data and test another hypothesis
You call a technician and he check with the car engine tries and finds out that the engine was out of order and needs repairing.
Draw conclusion:
The engine do not works when it is out of order and it is a cause of a car not being started.
<em>Now the theory and law making part can not be applied on this case but it is a part of scientific method.</em>
Hope it helps!
<span>anwser will be
F = ma
where
F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet
Substituting appropriately,
F = 0.005a --- call this Equation 1
Next working equation is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel
Substituting appropriately,
326^2 - 0 = 2(a)(0.83)
a = 64,022 m/sec^2
the anwser will be
Substituting this into Equation 1,
F = 0.005(64,022)
F =320.11 Newtons
Hope this helps. </span><span>
</span>
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = 
--------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = 
-------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v = 
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s
