A 0.025 kg bullet enters a 2.35-kg watermelon with a speed of 217 m/s and exits the opposite side with a speed of 109 m/s. If the melon was

originally at rest, then what speed will it have as the bullet leaves its

opposite side?

1.15m/s

Step-by-step explanation:

Given data

m1=0.025 kg

m2= 2.35kg

u1= 217 m/s

v1= 109 m/s

u2= 0

v2=???

Applying the law of conservation of linear momentum

m1u1+m2u2=m1v1+m2v2

substitute

0.025*217+2.35*0= 0.025*109+2.35*v2

5.425= 2.725+ 2.35v2

5.425-2.725=2.35v2

2.7=2.35v2

divide both sides by 2.35

v2= 2.7/2.35

v2=1.15 m/s

Hence the speed of the watermelon will be 1.15m/s

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

For every 8 cars there are 7 trucks

Therefore,

Cars:Truck=8:7

Answer is B)8:7

The solution is in the following image

The answer is in the image 

The answer is in the image 

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

The answer is in the image 

The solution is given in the image below