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liq [111]
2 years ago
7

In the winter, monarch butterflies travel from the United States to Mexico, where the weather is warmer. They return to the Unit

ed States in the spring.
-----------------------------------------------------------------------------
Which behavior does this scenario describe?

o Conditioning
o Hibernating
o Imprinting
o Migrating
Physics
2 answers:
Deffense [45]2 years ago
6 0
The answer is D: Migrating
Pavlova-9 [17]2 years ago
3 0

Answer: "Migrating", You're welcome

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Which statement is not true about radiation
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Which statement is correct? (2 points)
Lemur [1.5K]

Answer:

When a neutral atom looses an electron to another neutral atom, two charged atoms are created.

Explanation:

On the off chance that one of the two unbiased particle looses an electron, it turns out to be emphatically (charge: +1), in light of the fact that the electron conveys a charge of - 1. Thus, the other atom which acknowledges the electron turns out to be adversely charged (charge: - 1). And in the end, we will have two charged atoms.

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What is the specific heat capacity of water at 20 °C and 1 atmospheric pressure?
ycow [4]

Answer:

The specific heat of liquid water is about 4184 J/kg at 20 °C.

So, <u>Correct choice</u> - [C] 4186 J / kg ° C

6 0
3 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
Irina-Kira [14]

Answer:

part (a) \alpha\ =\ -2.5\ rad/s^2

part (b) N = 79.61 rev

part (c) \tau\ =\ 23.54\ Nm

Explanation:

Given,

  • Initial speed of the wheel = w_o\ =\ 50.0\ rad/s
  • total time taken = t = 20.0 sec

part (a)

Let \alpha be the angular acceleration of the wheel.

Wheel is finally at the rest. Hence the final angular speed of the wheel is 0.

\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

Let \theta be the total angular displacement of the wheel from initial position till the rest.

\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

We know,  1 revolution = 2\pi rad

Let N be the number of revolution covered by the wheel.

\therefore N\ =\ \dfrac{\theta}{2\pi}\\\Rightarrow N\ =\ \dfrac{500}{2\times 3.14}\\\Rightarrow N\ =\ 79.61\ rev

Hence the 79.61 revolution is covered by the wheel in the 20 sec.

part (c)

Given,

  • Mass of the pole = m = 4 kg
  • Length of the pole = L = 2.5 m
  • Angle of the pole with the horizontal axis = \theta\ =\ 60^o

Now the center of mass of the pole = d\ =\ \dfra{L}{2}\ =\ \dfrac{2.5}{2}\ =\ 1.25\ m

Weight component of the pole perpendicular to the center of mass = F\ =\ mgcos\theta

\therefore \tau\ =\ F\times d\\\Rightarrow \tau\ =\ 4\times 9.81\times cos60^o\times 1.25\\\Rightarrow \tau\ =\ 23.54\ Nm

3 0
3 years ago
A car of mass 1200 kg, moving with a speed of 72 mph on a highway,
Keith_Richards [23]

momentum(p) = mass(m) x velocity(v)

KE = kinetic energy = 1/2 mv²

a.

the ratio of the momentum of the SUV to that of the car

\tt \dfrac{mv}{1.5m\dfrac{2}{3}v }=1:1

b.  the ratio of the KE of the SUV to that of the car

\tt \dfrac{1/2\times mv^2}{1/2\times 1.5m\times (\dfrac{2}{3}v )^2}=\dfrac{mv^2}{1.5m\dfrac{4}{6}v^2 }=1:1

8 0
2 years ago
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