Two forces are acting on a 0.250 kg hockey puck as it slides along the ice. The first force has a magnitude of 0.340 N and point
1 answer:
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Answer:
When the ball is held motionless above the floor, the ball possesses only GPE energy.If the ball is dropped, its GPE energy decreases as it falls.If the ball is dropped, its KE energy increases as it falls.
Explanation:
If the ball is held motionless, then its kinetic energy is equal to zero, since kinetic energy depends on the velocity. And the ball is held above the ground, which means it possesses gravitational potential energy.
If the ball is dropped, its height will decrease, therefore its gravitational potential energy will decrease. Along the way, the ball will be in free fall, and therefore its velocity will increase, hence its kinetic energy.
Answer:
V = 5.83 m/s
Explanation:
Given that,
Mass of a ball of a clay, m = 2 kg
Initial speed of the clay, u = 35 m/s
Mass of a box, m' = 10 kg
Initially, the box was at rest, u' = 0
We need to find the velocity of the box after the collision. Let V be the common speed. Using the conservation of momentum to find it.
So, the velocity of the box after the collision is equal to 5.83 m/s.
Answer:
a) A1 =
b) A1 = 2.688 cm
c) Q1 = A1 x v1
d) v1 = 3.1994 m/s
e) A2 =
f) A2 = 0.7963cm
Explanation:
a) Area =
r =
thus,
area =
A1 =
A1 =
A1 = 2.688 cm
c) Flow rate = Area x velocity ( refer brainly.com/question/13997998)
Q1 = A1 x v1
d) From the above equation,
v1 = =
= 319.94 cm/s = 3.1994 m/s
e) Since the flow rate Q1 is constant throughout the hose, Av is a constant.
i.e. A1 x v1 = A2 x v2
thus,
A2 =
f) v2 = 10.8 m/s.
substituting the values in the above equation,
A2 = = 0.7963cm