Answer:
13.5
Explanation:
Mass: 5kg
Initial Velocity: -15
Final Velocity: 12
Force: 10
We can use the equation: Vf = Vi + at
We need to find acceleration, and we can use the equation, F=ma,
We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.
Now we have all the variables to find time.
Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)
Plugging them in into desmos gives 13.5 for time.
Ke= 1/2 x m x v^2
Ke= 1/2 x 2.1 x 30^2
Energy = 945 J
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.
Given Information:
Initial speed = u = 3.21 yards/s
Acceleration = α = 1.71 yards/s²
Final speed = v = 7.54 yards/s
Required Information:
Distance = s = ?
Answer:
Distance = s = 13.61
Explanation:
We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.
We know from the equations of motion,
v² = u² + 2αs
Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.
Re-arranging the above equation for distance yields,
2αs = v² - u²
s = (v² - u²)/2α
s = (7.54² - 3.21²)/2×1.71
s = 46.55/3.42
s = 13.61 yards
Therefore, the runner traveled a distance of 13.61 yards before being tackled.
Answer:
The person above me is right i had a test a couple of days ago and thats kinda what u put and got it right!
