Answer:
<h2>
<em>=</em><em> </em><em>2</em><em>4</em><em>5</em><em>.</em><em>2</em><em>5</em></h2>
Explanation:
<h3>150 x 7.5 x 9.81 =11036.25</h3><h3>ans / 60 =183.9</h3><h3>ans / 75 x 100 = 245.25</h3>
Answer:
t = 4.08 s
R = 40.8 m
Explanation:
The question is asking us to solve for the time of flight and the range of the rock.
Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:
We have these known variables:
- (v_0)_y = 0 m/s
- a_y = -9.8 m/s²
- Δx_y = -20 m
And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.
- -20 = 0t + 1/2(-9.8)t²
- -20 = 1/2(-9.8)t²
- -20 = -4.9t²
- t = 4.08 sec
The time it takes for the rock to reach the ground is 4.08 seconds.
Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.
List out known variables:
- v_0 = 10 m/s
- t = 4.08 s
- a_x = 0 m/s
We are trying to solve for:
By using the same equation, we can plug these known values into it and solve for Δx.
- Δx = 10 * 4.08 + 1/2(0)(4.08)²
- Δx = 10 * 4.08
- Δx = 40.8 m
The rock lands 40.8 m from the base of the cliff.
The total kinetic energy of the system after collision is 223.5J
In elastic collision kinetic energy and momentum are conserved.
According to the question
mass of object a = 25kg
mass of object b = 25 kg
initial velocity of a u1 = 5.98 m/s
initial velocity of b u2 = 0
so from momentum conservation-
mau1 + mbu2 = (m1+m2)v
25kg × 5.98m/s + 25×0 = (25+25)v
v = 2.99 m/s
Now the total kinetic energy after the collision will be:
final kinetic energy = 1/2 (m1+m2) v²
= 1/2 (25+25)× (2.99)²
= 223.5 J
Thus, total kinetic energy of the system after collision is 223.5J
Learn more about elastic collision here:
brainly.com/question/1808045
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<span>fission I think. I'm not 100% sure but I hope that helped you
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