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pogonyaev
3 years ago
13

Jeff is investigating factors that affect the growth rate of potted bean plants. Which of the following experimental variables w

ould not be relevant to his investigation?
A. the type of soil used
B. the amount of water given
C. the color of the pot
D. the amount of light exposure
Physics
2 answers:
Kipish [7]3 years ago
8 0
It would be C. the color of the pot. its pretty obvious that i would not effect the project.

aksik [14]3 years ago
5 0

Answer:

C. the color of the pot

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katovenus [111]

The formula that relates the 3 variables, namely: mass, volume, and density is: Density = mass / volume.

 

Let us identify each variable:

Density - more accurately called the volumetric mass density of a substance is its mass per unit volume. It is a key concept that examines how materials communicate in a lot of things, i.e weather, material sciences, engineering, and other fields of science.

Mass – defined as a property of a physical body in Physics, and when a net force is applied, it results  to the measure of a specific object’s resistance to acceleration.

 

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3 years ago
Calculate the impulse of a 0.140 kg baseball that starts at rest on a tee and reaches a final velocity of 20 m/s after being str
oee [108]

Explanation:

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What happens when a ray of light is directed at a mirror, a glass block and a prism?
zimovet [89]
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5 0
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How to get a + b on a graph
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The first positively essential requirement is that
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3 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
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