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maxonik [38]
3 years ago
12

Which scientific design has the fewest limitations?

Physics
1 answer:
AleksAgata [21]3 years ago
4 0
USING DISSECTION TO STUDY THE GENERAL STRUCTURE OF LILY FLOWERS.

Other choices are very hard to do. Maintaining a salt-water aquarium is very hard while going to an arctic tundra or space is impractical. 
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Give brief descriptions of both the Kuiper belt and the Oort cloud.
iren [92.7K]

Answer and Explanation:

KUIPER BELT : The Kuiper belt is on the name of the scientist kuiper .It is the largest structure of the solar system its inner edge is about 30 AU from the Sun. The thickness of the kuiper belt is about 10 AU. It contains comets and other icy bodies

OORT CLOUD : The name of this cloud is one the name of a scientist Jhon oort. It is a theoretical cloud and mainly consist icy bodies it far away from the Sun. Its structure is spherical in nature and mostly contain ice water amonia methane

3 0
4 years ago
What is the acceleration of a car that maintains a constant velocity of 100km/h for 10 s
Mademuasel [1]
A constant velocity means the acceleration is zero.
4 0
3 years ago
Read 2 more answers
A soccer goal is 2.44 m high. A player kicks the ball at a distance 13 m from the goal at an angle of 40°, and the ball just hit
Vera_Pavlovna [14]

Let <em>v</em> denote the initial speed of the ball. The ball's position at time <em>t</em> is given by the vector

\mathbf r(t)=v\cos40^\circ\,t\,\mathbf i+\left(v\sin40^\circ\,t-\dfrac g2t^2\right)\,\mathbf j

where <em>g</em> is the acceleration due to gravity with magnitude 9.80 m/s^2.

The ball reaches the goal 13 m away at time <em>t</em> such that

10\,\mathrm m=v\cos40^\circ t\implies t=\dfrac{10\,\mathrm m}{v\cos40^\circ}

at which point it attains a height of 2.44 m, so that

2.44\,\mathrm m=v\sin40^\circ\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)-\dfrac g2\left(\dfrac{10\,\mathrm m}{v\cos40^\circ}\right)^2

2.44\,\mathrm m=(10\,\mathrm m)\tan40^\circ-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)\left(\dfrac{100\,\mathrm m^2}{v^2\cos^240^\circ}\right)

\implies\boxed{v\approx3.75\dfrac{\rm m}{\rm s}}

6 0
4 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
NeX [460]

Answer:

The transverse displacement is   y(1.51 , 0.150) = 0.055 m    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    y(x,t) = Acos (kx -wt)

     The speed of the transverse wave is v = 8.25 \ m/s

     The amplitude of the transverse wave is A = 5.50 *10^{-2} m

     The wavelength of the transverse wave is \lambda = 0540 m

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          w = vk

Substituting values  

         w = 8.25 * 11.64

        w = 96.03 \ rad/s

The propagation constant k is mathematically represented as

                  k = \frac{2 \pi}{\lambda}

Substituting values

                  k = \frac{2 * 3.142}{0. 540}

                   k =11.64 m^{-1}

Substituting values into the equation for mechanical waves

      y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03  * 0.150))

       y(1.51 , 0.150) = 0.055 m    

4 0
4 years ago
the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. wh
Andrew [12]

Answer:

1750km

Explanation:

8 0
3 years ago
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