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slavikrds [6]
8 months ago
10

which of the following safety harnesses is the safest and most recommended when hunting from a treestand?

Physics
1 answer:
Lyrx [107]8 months ago
4 0

The best tree stand safety harness is the Hunter Safety System Hybrid Flex Safety Harness, with its awesome ElimiShield Scent Control Technology.

Moreover, These stands are designed to be attached directly to the tree. Hunters using a fixed or suspended stand must choose a method of climbing up and down from the platform. The safest and most used method is sectional ladders.

You can learn more about this at:

brainly.com/question/28335498#SPJ4

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an object of mass 6000 kg rests on the flatbed of a truck. it is held in place by metal brackets that can exert a maximum horizo
Otrada [13]

Answer:

minimum stopping distance will be d = 75 m

Explanation:

Maximum force exerted by the bracket is given as

F = 9000 N

now we know that mass of the object is

m = 6000 kg

so the maximum acceleration that truck can have is given as

F = ma

9000 = 6000 a

a = 1.5 m/s^2

now for finding minimum stopping distance of the truck

v_f^2 - v_i^2 = 2a d

0^2 - 15^2 = 2(-1.5) d

d = 75 m

4 0
3 years ago
Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)
SpyIntel [72]

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

5 0
2 years ago
Which are characteristics of light? Check all that apply
Schach [20]

Answer:

wavelength, frequency, energy and speed

Explanation:

7 0
2 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0
2 years ago
The maximum value of magnetic in an electric field 3.2 *10^4​
Svetradugi [14.3K]

Answer:

the answer is 12 because if your magnetic value and Electric field is 3.2 the answer will be 12

6 0
2 years ago
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