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Rina8888 [55]
3 years ago
6

Which of the following describes qualitative data?

Physics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

The answer is C

Explanation:

All other have definitive measurements of physical quantities that is fixed for everybody under given conditions whereas the C can vary upon person to person as per their observation.

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Which group of animals would be served best by the following adaptations?
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The answer to this question would be: <span>A) animals that live in deserts 

</span>Desert temperature is high, especially in the day, <span>An animal that lives in the desert needs to adapt to the high temperature either by reducing the heat or by increasing heat loss. By becoming nocturnal, the animal also able to evade the sunlight so it was less exposed to the heat. 
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Even if we do not send people to far-flung planets, we need to send some information about ourselves and our planet.
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Lifting a stone block 146m to the top of the Great Pyramid required 146,000 J of work. How much work was done to lift the block
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B is your answer to the question
3 0
3 years ago
A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,
muminat

Answer:

The speed of the plank relative to the ice is:

v_{p}=-0.33\: m/s

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

m_{g}v_{g}+m_{p}v_{p}=0 (1)

Where:

  • m(g) is the mass of the girl
  • m(p) is the mass of the plank
  • v(g) is the speed of the girl
  • v(p) is the speed of the plank

Now, as we have relative velocities, we have:

v_{g/b}=v_{g}-v_{p}=1.55 \: m/s (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

45v_{g}+168v_{p}=0

v_{g}-v_{p}=1.55

v_{p}=-0.33\: m/s

I hope it helps you!      

8 0
3 years ago
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
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