What is the name for the narrowing at the muzzle end of a shotgun barrel?

Answers

Answer 1
Answer:

The pellets in a shotshell start to spread or scatter as soon as they are fired. This dispersion will increase as the pellets travel farther. Shotgun barrels feature a choke to regulate this pellet pattern. The shotgun barrel's choke is the constriction at the muzzle end.

How long is a shot string?

At 20 yards, it can be roughly 5 feet long, and at 30 yards, it can be 8 feet long. The variance is brought on by the pellets' distortion as they move along the barrel.

The full choke features the tightest shot pattern and is the tightest choke. The pellets are kept together for a longer distance thanks to this choke. It is typically used for shots that are further than 40 yards.

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Answer 2
Answer: The name for the narrowing at the muzzle end of a shotgun barrel is the choke.

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A police car is parked by the side of a highway and its siren emits a sound measuring 750 Hz. As you approach the police car from a distance, you hear a change in the sound. The change is in the sound's

Answers

As you approach the police car with blaring siren, you hear a change in pitch and frequency. The apparent pitch changes as a result of the Doppler effect. Frequency determines pitch; frequency and wavelength are inversely proportional. You wouldn't "hear" either one. The result is pitch.

D) pitch and frequency.

Answer:

The correct answer would be frequency, intensity, and amplitude.

It can be explained with the help of Doppler effects which states that the frequency or wavelength of sound changes (increases or decreases) as source and observer move towards or away from each other.

Due to this effect, the frequency and intensity of siren increase as we move towards the siren.

In addition, amplitude also increases as we move towards the source of the sound and decreases as the observer moves away from the source.

The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.. . Part C. . What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.. . A) 0, 0, 0, 0. B) 0, 0, 0, -9.80. C) 15.0, 0, 0, 0. D) 15.0, 0, 0, -9.80. E) 0, 26.0, 0, 0. F) 0, 26.0, 0, -9.80. G) 15.0, 26.0, 0, 0. H) 15.0, 26.0, 0, -9.80

Answers

Final answer:

At the peak of its trajectory, the ball's vertical velocity (v1,y) is zero, whereas its horizontal velocity (v1,x) maintains its consistency. The accelerations a1,x and a1,y are 0 and -9.8 m/s² respectively.

Explanation:

The physical concept analyzed here is projectile motion. Considering the trajectory of a ball, we can separate its motion into horizontal and vertical components. At the peak of its trajectory (time t1), the ball's vertical velocity (v1,y) is zero, because it temporarily stops moving upwards before it begins to fall again. However, since there is no acceleration in the horizontal direction (a1,x), the ball keeps moving horizontally with constant velocity (v1,x).

The acceleration in the vertical direction (a1,y) is still the acceleration due to gravity (-9.8 m/s²). This is because even at the peak of the trajectory, the ball is still accelerating downwards. Therefore, the correct answer should include v1,y = 0 m/s, v1,x = consistent value, a1,x = 0 m/s², and a1,y = -9.8 m/s². So the most probable answer is D) 15.0, 0, 0, -9.80, assuming the ball's horizontal velocity is 15 m/s.  Please check the exact value of horizontal velocity (v1,x) in the settings of your problem.

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Displacement vectors A, B, and C add up to a total of zero. Vector A has a magnitude of 1550 m and a direction of 25.6° north of east. Vector B has a direction of 41.0° east of south, and vector C has a direction of 35.1° north of west. Find the magnitudes of vector B and vector C.

Answers

Answer:

B= 5626.77 m

C= 6220. 5 m

Explanation:

Because the sum of the vectors must be equal to zero, then the result  force in x and the result force  in y must be  zero.

We propose 2 equations  x-y to solve the problem:

Rx :resulting from forces at x

Ry: resulting from forces at y

Rx= Ax+Bx+Cx=0

Ry= Ay+By+Cy=0

Ax =1550 *cos25.6°= 1397.84

Ay =1550 *sin25.6° = 669.73

Bx= B*sin41° = 0.656B

By= -B*cos41° = -0.7547 B

Cx= -C*cos35.1°= -0.8181 C

Cy= C* sin35.1° = 0.575 C

Rx= 1397.84+0.656B-0.8181 C=0

Ry= 669.73-0.7547 B+ 0.575 C=0

System of 2 equations with 2 incognites:

+0.656B-0.8181 C= - 1397.84

-0.7547 B+ 0.575 C= -669.73

Resolving the system:

B= 5626.77 m

C= 6220. 5 m

Final answer:

Since the vectors A, B, and C add to zero, their x and y components should add up to zero as well. This information helps create a system of linear equations to solve for the magnitudes of vectors B and C.

Explanation:

The subject matter of this vector question involves physics, specifically vector addition and vector resolution. The key point you must understand here is that these vectors add up to a total of zero. This gives us an equation to solve for the magnitudes of the other vectors, B and C.

First, break each vector into its x (east-west) and y (north-south) components and set up the system of equations. For example, Ax = 1550 cos(25.6°), Ay = 1550 sin(25.6°). Notice that the direction for vector B is given as east of south, which means B's x component is negative and its y component is positive.

When these are added together they should equal zero (since A + B + C = 0), so you will have two equations to solve for the magnitudes of B and C. This is a system of linear equations you'll need to solve, either manually or with calculator.

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The scattering of a stream of positively charged particles when striking a thin film of gold confirms thatthe positive charge in an atom must be located in a dense central mass.atoms contain electrons in varying energy layers around the nucleus.

electrons are found in the center of an atom.

electrons can only be found within predicted areas surrounding the nucleus.

Answers

Answer: The correct statement is the positive charge in an atom must be located in a dense central mass.

Explanation:

Rutherford gave an experiment called as gold foil experiment.

In his experiment, he took a gold foil and bombarded it with alpha particles which carry positive charge of +2 units. He at first thought that the particles will pass straight through the foil, but to his surprise, many of them passed through, some of them deflected their path and a few of them bounced back straight from their path.

From this observations, he concluded that in an atom, there exist a small positive charge that is present at the center. Because of this positive charge, the alpha particles deflected their path and some of them bounced straight back their path.

Thus, the correct statement is the positive charge in an atom must be located in a dense central mass.

the positive charge in an atom must be located in a dense central mass

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force constant of 8.08 N/m. When the cannon is fired, the ball moves 15.8 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.031 0 N on the ball.With what speed does the projectile leave the barrel of the cannon?

Answers

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=(1)/(2)kx^2=(1)/(2)* 8.08* 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031* 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

(1)/(2)mv^2=0.0052

(1)/(2)* 0.00524* v^2=0.0052

v = 1.40 m/sec

An atom of carbon (C) forms covalent bonds with two atoms of oxygen (O) to form carbon dioxide. How are the valence electrons of these atoms rearranged to form the bonds?A. Electrons are transferred from the carbon atom to the oxygen atoms.
B. Electrons are transferred from the oxygen atoms to the carbon atoms.
C. Many valence electrons are shared between the atoms.
D. A few valence electrons are shared between the atoms

Answers

Answer : The correct option is C.

Explanation :

Covalent bond : Covalent bond forms when two non-metal atom share a pair of electrons.

In the covalent bond formation of carbon dioxide, carbon and oxygen atom share four pair of electrons. The carbon atom joined by four covalent bonds to two oxygen atoms.

The valence electrons in carbon atom = 4

The valence electrons in two oxygen atom = 2(6) = 12

Therefore, these atoms are rearranged to form the bonds by the sharing of many valence electrons between the atoms.

The lewis-dot structure of carbon dioxide is shown below.

Answer:

Person above me got it correct

Explanation:

The answer is C Many valence electrons are shared between the atoms

Just confirming :)