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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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It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0
3 years ago
What’s is a <br> genotype
Firdavs [7]

Answer:

The genetic makeup in a organism.

Explanation:

8 0
3 years ago
Read 2 more answers
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tis
Eduardwww [97]

Answer:

m' = 1 x 10⁻⁶ kg/s

Explanation:

Given that

Diffussion constant = 1 x 10⁻¹¹

Mass flow rate ,m = 2 x 10⁻⁶ kg/s

The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.

So new flow rate m'

m'=\dfrac{m}{2}

m'=\dfrac{2\times 10^{-6}}{2}\ kg/s

m' = 1 x 10⁻⁶ kg/s

6 0
3 years ago
A horizontal uniform board of weight 125N and length 4 m is supported by vertical chains at each end. A person weighing 500N sit
Misha Larkins [42]

Let's apply an equation of equilibrium to the situation: The sum of the moments about the left end of the board must equal 0.

We have three moments to add. Positive force values indicate upward direction and negative values indicate downward direction. All distances given below are measured to the right side of the left end of the board:

  1. The weight of the board, -125N, located at 2m (center of the board due to its uniform density)
  2. The tension in the right chain, +250N, located at 4m
  3. The weight of the person, -500N, located at a distance "x"

The sum of the moments must equal 0 and is given by:

ΣFx = 0

F is the magnitude of force, x = distance from the left end of the board

Plug in all of the force and distance values and solve for x:

ΣFx = 250(4) - 125(2) - 500x = 0

500x = 750

x = 1.5m

7 0
3 years ago
A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra
monitta
Your answer to your question is 6.9 m/s
5 0
3 years ago
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