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Kay [80]
3 years ago
7

Young animals often learn valuable hunting skills through

Physics
2 answers:
pogonyaev3 years ago
8 0
Young animals often learn valuable hunting skills through playing/wrestling.
Hope it helps!
uysha [10]3 years ago
7 0
They're Farther's, Genes.
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What would be the most accurate way to describe resistance?
ohaa [14]

Explanation:

hit is the abilty to accept the flow of current and is measured in ohoms

7 0
3 years ago
Read 2 more answers
A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car
dangina [55]
Her magnitude of deceleration on the ice would be 15.126m/s
5 0
2 years ago
What are the concepts of dynamically continuous innovation and discontinuous innovation? Can you share examples to illustrate th
KiRa [710]

Answer:

Explained

Explanation:

Dynamically continuous innovation:

- Falls in between continuous and discontinuous innovation.

-Changes in customer habits are not as large as in discontinuous innovation and not as negligeble as in continuous innovation.

best example can as simple as transformation in  Television. New HD TVs have flat panels, wide screens and improved performance The Added features are considered dynamically improved.

Discontinuous innovation:

- discontinuous innovation comprise of new to world product only so they are discontinuous to every customer segment.

- these product are so fundamentally different from the the product that already exist that they reshape market and competition.

For example- the mobile and the internet technology are reshaping the market through regular innovation and change.

7 0
3 years ago
Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre
Elanso [62]

Answer:

d= 0.242 mm

Explanation:

Slit width (d ) = ?

Screen distance ( D ) = 1.25 m

Wave length of light λ = 600 nm

Distance of n the dark fringe from centre

= n λ D / d

Here n = 2

so

6.2\times10^{-3}=\frac{2\times600\times10^{-9}}{d}

d=\frac{1500\times10^{-6}}{6.2}

d= 0.242 mm

4 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
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