Answer:
 ms⁻¹
 ms⁻¹
Explanation:
Consider the motion of the bullet-block combination after collision
 = mass of the bullet = 0.0382 kg
 = mass of the bullet = 0.0382 kg 
 = mass of wooden block = 3.78 kg
 = mass of wooden block = 3.78 kg 
 = velocity of the bullet-block combination after collision
 = velocity of the bullet-block combination after collision
 = spring constant of the spring = 833 N m⁻¹
 = spring constant of the spring = 833 N m⁻¹
 = Amplitude of oscillation = 0.190 m
 = Amplitude of oscillation = 0.190 m
Using conservation of energy 
Kinetic energy of  bullet-block combination after collision = Spring potential energy gained due to compression of spring 


 ms⁻¹
 ms⁻¹
 = initial velocity of the bullet before striking the block
 = initial velocity of the bullet before striking the block
Using conservation of momentum for the collision between bullet and block


 ms⁻¹
 ms⁻¹
 
        
             
        
        
        
Answer:
t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)
S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream
The problem does not specify V the speed of the river
 
        
                    
             
        
        
        
The answer is B. 
She can measure the mass of the water, marble and the graduated cylinder with the balance.
The volume of the water can be shown on the marked graduated cylinder, the volume of the marble can be measured by the volume difference of the water before and after the marble is put in.
        
                    
             
        
        
        
Answer: Impulse = 4 kgm/s 
Explanation:
From the question, you're given the following parameters:
Momentum P1 = 12 kgm/s 
Momentum P2 = 16 kgm/s 
Time t = 0.2 s
According to second law of motion, 
Force F = change in momentum ÷ time
That is 
 F = (P2 - P1)/t
Cross multiply 
Ft = P2 - P1
Where Ft = impulse 
Substitute P1 and P2 into the formula 
Impulse = 16 - 12 = 4 kgm/s 
The magnitude of the impulse is therefore 4 kgm/s. 
 
        
             
        
        
        
The weight of the car in the picture of the computer screen is 9,800 Newton's.