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disa [49]
3 years ago
15

Tell me about an adaptation that you know of that is a survival adaptation

Physics
2 answers:
BigorU [14]3 years ago
8 0
Camouflage because they blend in
Fed [463]3 years ago
4 0
Zebra coloration - the coloration makes it harder for predators to spot.
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Which of the following should be included in a family communication plan? A. A list of favorite foods and beverages B. A meeting
Ivanshal [37]

Answer:

I believe it's B. A meeting spot for family members

Explanation:

Make copies of your Family Emergency Communication Plan for each member of the household to carry in his or her wallet, backpack, or purse. So they know where to go in case of an emergency!

5 0
3 years ago
Read 2 more answers
A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
lisabon 2012 [21]

Answer:

\alpha =10.93radian/sec^2

Explanation:

We have given given the final angular velocity \omega _{final}=13.5rad/sec

And \omega _{initial}=22rad/sec

Displacement \Theta =13.8radian

We have to find the angular acceleration \alpha

According to law of motion \omega _{final}^2=\omega _{initial}^2+2\alpha \Theta

So 13.5^2=22^2+2\times \alpha \times 13.8

\alpha =-10.93radian/sec^2

In question we have tell about magnitude only so \alpha =10.93radian/sec^2

4 0
3 years ago
A firefighter directs a stream of water from a fire hose at an angle of 40.0° above the horizontal. If the velocity of the strea
mina [271]

Answer: The time of motion of the stream of water is calculated as follows;

X = Vₓt

Explanation:

5 0
2 years ago
A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou
ELEN [110]
This problem can be solved using a kinematic equation. For this case, the following equation is useful:

v_final = v_initial + at

where,
v_final = final velocity of the nail
v_initial = initial velocity of the nail
a = acceleration due to gravity = 9.8 m/s^2
t = time 

First, we determine the time it takes for the nail to hit the ground. We know that the initial velocity is 0 m/s since the nail was only dropped. It has a final velocity of 26 m/s. We substitute these values to the equation and solve for t:

26 = 0 + 9.8*t
t = 26/9.8 = 2.6531 s

The problem asks the velocity of the nail at t = 1 second. We then subtract 1 second from the total time 2.6531 with v_final as unknown.

v_final = 0 + 9.8(2.6531-1) = 16.2004 m/s.

Thus, the nail was traveling at a speed of 16. 2004 m/s, 1 second before it hit the ground. 

5 0
4 years ago
A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric
Tcecarenko [31]

Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

8 0
4 years ago
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