A 4 kg bowling boll sliding to the right at 8 m/s has elastic head-on collision with another 4K bowling ball initially at rest.
1 answer:
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Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector
y component of the vector
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector
y component of the vector
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where =Mass of block B
Substitute the values
For block A
Where Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N