Although many characteristics are common<span> throughout the </span>group<span>, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the </span>Group<span> 1 Alkali Metals. All the </span>elements<span> in </span>Group 2 have two<span> electrons in their valence shells, giving them an oxidation state of +</span><span>2.</span>
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Answer:
The canon B hits the ground fast.
Explanation:
Given that,
Speed of cannon A = 85 m/s
Speed of cannon B= 100 m/s
Speed of cannon C = 75 m/s
We need to calculate the cannonballs will hit the ground with the greatest speed
Using conservation of energy
The final kinetic energy of canon depends on initial kinetic energy and potential energy.
The final velocity depends upon initial velocity and initial height.
So, the initial velocity of canon B is high.
Hence, The canon B hits the ground fast.
Answer:
Explanation:
Given
mass of book(m)=2.1 kg
height up to which book is lifted is (h)2.2 m
height of person 
Potential energy of book relative to ground=mgh

(b)PE w.r.t to person head =mg(h-h0)

work done by person in lifting box 2.2 m w.r.t floor
Word done =Potential Energy of box relative to floor=45.2 J
Answer:
PE=0.29J
Explanation:
According to the description, there is a angle and in point swung upward of 70°
So,

Appling the equation of Potential Energy we have,
