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Delicious77 [7]
3 years ago
9

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

s to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x2, where x is in meters.
Physics
1 answer:
Mashutka [201]3 years ago
5 0

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

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Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.
Oxana [17]
I think the answer is “greenhouse effect”
5 0
4 years ago
Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?
Alexandra [31]

11) 1.04\cdot 10^7 J

12) 1.04\cdot 10^7 J

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

PE=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

g=9.8 m/s^2 (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

PE=(8325)(9.8)(127)=1.04\cdot 10^7 J

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

KE_t + PE_t = KE_b + PE_b

where

KE_t is the kinetic energy at the top

PE_t is the potential energy at the top

KE_b is the kinetic energy at the bottom

PE_b is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

KE_t=0

Also, at the bottom the height is zero, so the potential energy is zero

PE_b=0

Therefore, we find:

KE_b=PE_t=1.04\cdot 10^7 J

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

KE=1.04\cdot 10^7 J

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

KE_1 + PE_1 = KE_2 + PE_2

where

KE_1 is the kinetic energy at the top of the 1st hill

PE_1 is the potential energy at the top of the 1st hill

KE_2 is the kinetic energy at the top of the 2nd hill

PE_2 is the potential energy at the top of the 2nd hill

From the previous questions, we know that

KE_1=0

and

PE_1=1.04\cdot 10^7 J

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J

So, the kinetic energy at the second hill is

KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J

And so, the speed is

v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s

4 0
3 years ago
develop a plan to test the hardness of a sample of feldspar using the following items glass plate copper penny and streak plate
Artemon [7]
Stroke then 3883 you choke <3333333333
6 0
3 years ago
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

4 0
3 years ago
Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Zielflug [23.3K]

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

4 0
3 years ago
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