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STALIN [3.7K]
2 years ago
12

What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o

f a rifle?
Physics
1 answer:
leonid [27]2 years ago
8 0

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,  

d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}

This value can be used to determine the time for the pellet through the barrel:

t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s

Finally, we can use the above to calculate the force:

F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N



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An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
A car travels 1008km in 14 hours. What is the average speed of the car?
Montano1993 [528]

Answer:

72 km/hr. ...jshdhdhddjidididdiididudd

6 0
3 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

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2 years ago
In 10 seconds a car accelerates 4/ms to 60m/s. how fast was the car going before it accelerated?
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Well it seems like this problem gives you what you need.  You said the car was going 4m/s and then accelerated to 60m/s... so 4m/s would be your answer for the initial velocity
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Two loudspeakers emit identical sound waves along the x-axis. The sound at a point on the axis has maximum intensity when the sp
EleoNora [17]

The concept required to solve this problem is related to the wavelength.

The wavelength can be defined as the distance between two positive crests of a wave.

The waves are in phase, then the first distance is

\Delta x_1 = 20cm

And out of the phase when

\Delta x_2 = 30cm

Thus the wavelength is

\Delta x_2-\Delta x_1 = \frac{\lambda}{2}

Here,

\lambda =  Wavelength

If we rearrange the equation to find it, we will have

\lambda = 2 (\Delta x_2-\Delta x_1 )

\lambda= 2(30-20)

\lambda = 20cm

Therefore the wavelength of the sound is 20cm.

5 0
3 years ago
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