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vlabodo [156]
3 years ago
13

What distance should be used to pattern a shotgun hunter ed?

Physics
1 answer:
Murljashka [212]3 years ago
7 0
I think the distance that should be used is the distance that one expects to be from the game you are hunting. Before taking a shotgun for a gobbler or even for ducks or other animals, you need to see how your gun performs by patterning it at various ranges with the load you want to use.
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A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

(c) energy is being delivered by the battery?

The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

4 0
3 years ago
How can scientific phenomena be used in design?​
Alex73 [517]
I’m not really sure I’m sorry
4 0
3 years ago
An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0
3 years ago
(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
3 years ago
1. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to Newton's Third Law?  Action Force:
Natali [406]
The Action Force of this scenario is the pushing force of the Diver. The Reaction Force is the raft pushing back on the diver. 

The Third Law of Motion states that "For every action, there is an equal and opposite reaction." Now when the diver dives off the raft, the raft is also pushing the same amount of force as the diver did as he dives off. The diver will then move forward and the raft on the other hand will move backwards.

The movement of the raft shows the opposite force. It will move backwards depending on how strong the diver will push off on the raft. And the amount of force he pushes on it, the raft will exert the same force so the stronger the force of the diver, the farther he will go because the raft will push him in that same direction as it goes backwards. 
7 0
3 years ago
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