Data:
The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 

<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 
Q = 4 </span>μC → 



<span>
We have:
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Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum 
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v 
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2 
                                                = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
                                                = 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2                 
                                                 = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J 
 
        
             
        
        
        
The correct answer for this question is this one: "The drops dripped from a bloody knife about 2 ft above the ground."
<span>On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. The statement that best accounts for the drops is that <em>the </em></span><span><em>drops dripped from a bloody knife about 2 ft above the ground.</em>
</span>
Hope this helps answer your question and have a nice day ahead.
        
             
        
        
        
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
<h3>
Frictional force between the block and the horizontal surface</h3>
The frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Learn more about frictional force here: brainly.com/question/4618599
 
        
             
        
        
        
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg