1.) Two positive charges of 4.3 mC each are separated by 0.25 m.
 What is the size and type of force between the two charges? Help
2.) A positive charge of 1.32 x 10-4 C and a negative charge of 9.8 mC
 are 0.015 m apart. What is the size and type of force between them? H
3.) Two electrons in an atom are separated by 1.5 x 10-10 m , the typical size
 of an atom. What is the size and type of force between them? H
4.) A negative charge of 4.7 mC exerts a repulsive force of 51.0 N on a 2nd charge
 0.062 m away. What is the size and polarity (pos or neg) of the 2nd charge? H
5.) A negative charge of 9.3 x 10-5 C exerts an attractive force of 37.4 N when
 placed 0.080 m away from a 2nd charge. Find size and polarity of 2nd charge. H
6.) Two positive charges of 3.0 mC exert a repulsive force of 2.0 N on each other. 
 By what distance are they separated? H
7.) Two charges q1 and q2 are separated by a distance, d and exert a force, F1
 on each other. What new force (F2) will exist if:
 a) q1 is doubled. H d) d is doubled. H 
 b) q1 is doubled and q2 is tripled. H e) d is tripled. H 
 c) q1 is cut in half. H f) q2 is doubled and d is cut in half. H
8.) a) How many electrons are there in 1 C ? H
 b) How many excess electrons are on a ball with a charge of -5.26 x 10-17 C ? H
 c) How many excess protons are on a ball with a charge of 7.29 x 10-12 C ? H
9.) How many coulombs of negative charge does a 5 gram nickel coin have?
 [takes 3 steps; do a) , b) and c) to get answer, don't round off until part c]
 a) Find the number of atoms in a nickel.
 Its mass is 5 g and nickel has 6.02 x 1023 atoms / 58 g. H
 b) Find the number of electrons in the coin. There are 28 electrons / atom. H
 c) Find how many coulombs of negative charge in a nickel. 1.6 x 10-19 C / 1 e- H
10.) A lightning bolt transfers 35 C to Earth. How many electrons are transferred? H
        
             
        
        
        
Note: The answer choices are :
a) Increased
b) Decreased
c) stayed the same
Answer:
The correct option is Increased
The magnitude of the electric field potential difference between the wingtips increases.
Explanation:
The magnitude of the electric potential difference is the induced emf and is given by the equation:

where l = length
v = velocity
B = magnetic field
As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e.  ,
 , 
the induced emf = vlB, and thus increases.
The magnitude of the electric field potential difference between the wingtips increases
 
        
             
        
        
        
The answer is going to be element #29 Copper makes blue 
Red:#38 
Green:#56
Pink:#3
Yellow:#11
Gold:#20
Hopes This Helps 
        
                    
             
        
        
        
Answer:
(a) m = 1.6 x 10²¹ kg
(b) K.E = 2.536 x 10¹¹ J
(c) v = 7.12 x 10⁵ m/s 
Explanation:
(a)
First we find the volume of the continent:
V = L*W*H
where,
V = Volume  of Slab = ?
L = Length of Slab = 4450 km = 4.45 x 10⁶ m
W = Width of Slab = 4450 km = 4.45 x 10⁶ m
H = Height of Slab = 31 km = 3.1 x 10⁴ m
Therefore,
V = (4.45 x 10⁶ m)(4.45 x 10⁶ m)(3.1 x 10⁴ m)
V = 6.138 x 10¹⁷ m³
Now, we find the mass:
m = density*V
m = (2620 kg/m³)(6.138 x 10¹⁷ m³)
<u>m = 1.6 x 10²¹ kg</u>
<u></u>
(b)
The kinetic energy will be:
K.E = (1/2)mv²
where,
v = speed = (1 cm/year)(0.01 m/1 cm)(1 year/365 days)(1 day/24 h)(1 h/3600 s)
v = 3.17 x 10⁻¹⁰ m/s
Therefore,
K.E = (1/2)(1.6 x 10²¹ kg)(3.17 x 10⁻¹⁰ m/s)²
<u>K.E = 2.536 x 10¹¹ J</u>
<u></u>
(c)
For the same kinetic energy but mass = 77 kg:
K.E = (1/2)mv²
2.536 x 10¹¹ J = (1/2)(77 kg)v²
v = √(2)(2.536 x 10¹¹ J)
<u>v = 7.12 x 10⁵ m/s</u>
 
        
             
        
        
        
Answer:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m 
height of basket is 3.05 m
Launching angle

y=1.05
equation of trajectory of ball is given by

for x=12.27

u=10.69
for x=11.73

u=11.436 m/s
Thus range of speed is (10.69, 11.436)