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Scorpion4ik [409]
3 years ago
8

Explain your answers to 9a and 9b in terms of Newton's laws of motion.

Physics
1 answer:
kogti [31]3 years ago
4 0

Answer:i One way to solve the quadratic equation x2  = 9 is to subtract 9 from both sides to get one side equal to 0: x2  – 9 = 0. The expression on the left can be factored:

Explanation:

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It would have to be c because it is a chemical change. your welcome!!!!!
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Margaret is the new director of research at a well-known pharmaceutical company. She has been asked to design a set of research
Serjik [45]

Answer:

Double blind experiment

Explanation:

It is an experimental method , which helps to avoid any impartiality and any error due to biasing .

The experiment give rise to very accurate results , which is very important for any experiment .

Hence , the new director Margaret , need to design a set of double - blind experiments.

7 0
3 years ago
30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
Aleks04 [339]

Answer:

I think it's the most important part in this

7 0
3 years ago
A 1502.7 kg car is traveling at 33.1 m/s when
Bond [772]

By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0
3 years ago
A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
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