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Tanya [424]
4 years ago
6

A 10.1 g bullet leaves the muzzle of a rifle with a speed of 558.4 m/s. what constant force is exerted on the bullet while it is

traveling down the 0.7 m length of the barrel of the rifle? answer in units of n
Physics
1 answer:
artcher [175]4 years ago
6 0
To answer any question to do with mass, acceleration, velocity (etc...) you use one or more of this set of equations. They are called the 'Kinematic' equations and are really easy to re-arrange, when you need to find the 'unknown quantity' that the question is asking for. Below these equations i will show you the general problem solving strategy to very quickly and easily solve these problems.

The Kinematic Equations

D = (v_{i} * t) + (0.5 * a * t^2)
v_{f} ^2 = v_{i}^2 + (2 * a * D)
v_{f} = v_{i} + (a*t)
D =  (\frac{v_{i} + v_{f}}{2} ) * t

The Method

1. 
Write down all the known variables and their values + the one you don't know with a question mark next to it. (v = ?)

2. Find the kinematic equation where you have ALL the variables <span>except</span> the one you are missing.

3. Re-arrange the equation to solve for the missing variable.

Done!

<span>HINT: Sometimes you need to use two kinematic equations, one after the other, in order to get the answer. It is common for questions to not give two variables, like velocity AND acceleration to force you to solve for one, then use that solution in another kinematic equation to finally solve the problem. Writing down the known variables is the fastest way to see what you have and what you need.
____________________________________________

</span>This question takes things one step further and is asking you to find the FORCE on the bullet. To convert to a force in any situation you need to find the <span>acceleration</span> and multiply it by the mass as follows:

F = m * a

<em>Remember this equation!! you will use it ALL the time.

</em>Please let me know if you still cannot solve this problem, but i believe you should be fine :)<em>
</em>
You might be interested in
How would you describe brass since it ia used in weapons, pipes, intruments and ect? A compound, Alloy, Element l, Or molecule?
Alik [6]
<span>Brass is an <u>alloy</u>. An alloy is a mixture of elements to form a unique material. Brass is a mixture of copper and zinc and the percentage of each element depends on the desired material. It has a higher malleability than bronze or zinc. Meaning that it can be bend easily into it desired form.</span>
5 0
3 years ago
Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The
Kay [80]

Answer:

25.33 rpm

Explanation:

M = 100 kg

m1 = 22 kg

m2 = 28 kg

m3 = 33 kg

r = 1.60 m

f = 20 rpm

Let the new angular speed in rpm is f'.

According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.

Initial angular momentum = final angular momentum

(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =

                                  (1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'

(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'

( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'

2660 = 105 x f'

f' = 25.33 rpm

8 0
3 years ago
A person jogs 4.0 km in 32 minutes, then 2.0 km in 22 minutes, and finally 1.0 km in 16 minutes. What is the jogger's average sp
kupik [55]

Answer:

22

Explanation:

i did it before

7 0
4 years ago
a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

5 0
4 years ago
Use this media to help you complete the question.
Papessa [141]

Answer:

589.3 g is the answer

Explanation:

6 0
3 years ago
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