In space (no gravity or friction), you throw a ball with mass 0.1 kg at a target with mass 1 kg. You throw the ball at a speed o
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Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Answer:
the position of the wood below the interface of the two liquids is 2.39 cm.
Explanation:
Given;
density of oil, = 926 kg/m³
density of the wood, = 974 kg/m³
density of water, = 1000 kg/m³
height of the wood, h = 3.69 cm
Based on the density of the wood, it will position across the two liquids.
let the position of the wood below the interface of the two liquids = x
Let the wood be in equilibrium position;
Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.