A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. Exactly 5.0 s later, its speed is 6.0 m/s. What is the puck’s average acceleration, and what is the coefficient of kinetic friction between puck and ice?
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Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting was
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumg
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force is
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force isf/m=mug
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force isf/m=mugAnd this is equal to a_v. So,
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force isf/m=mugAnd this is equal to a_v. So,a_v=mug
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force isf/m=mugAnd this is equal to a_v. So,a_v=mugPutting g=9.8ms^-2, we get
Answer:Average acceleration is -"1.2 ms"^-2 and coefficient of frictional force is 0.12Explanation:Average deceleration (a_v) is equal to change in velocity in given time.So, a_v =((12-6) \ "ms"^(-1))/"5 s" = 6/5 \ "ms"^-2Now, if this happened due to frictional force then we can say that the frictional force acting wasf=muN=mumgwhere mu is coefficient of frictional force and m is its mass.So, the deceleration caused by the force isf/m=mugAnd this is equal to a_v. So,a_v=mugPutting g=9.8ms^-2, we getmu=a/g=(6/5) \ "ms"^(-2)xx(1/(9.8 \ "ms"^(-2)))=0.12
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