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Mazyrski [523]
4 years ago
9

What is electropositivity and electronegativity? Between group 1 and group 17, which is electropositive and electronegative?

Physics
1 answer:
marishachu [46]4 years ago
3 0
Electronegativity is a chemical property that describes the tendency of an atom or a functional group to attract electrons towards the nucleus. Electropositivity on the other hand is <span>a measure of an element's ability to donate electrons. Group 1 is a group of metals hence they are electropositive while group 17 is electronegative</span>
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A rocket is dropped out of an airplane at 100 m/s (downward). If the rocket fires causing an upward acceleration of Ct2 and it t
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Answer:

C = 0.0125 m/s⁴. The calculation procedure can be found in the attachment below. The concept of motion along a straight line with constant acceleration has been applied to solve the problem.

Explanation:

The sign convention chosen in this problem solution is upwards as positive and downwards negative. The equation of motion v = u + at has been used to calculate the constant C as only one unknown is contained in this equation. This is so because we have been given the initial velocity, the acceleration and the time taken. To solve future problems of this kind, first thing to check for is an equation of motion with the least number of unknown. This helps to reduce the complexity of the problem solution.  

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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

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