Answer:
1 kg lead to earth is greater attraction as mass of earth is much more than 1kg lead.
Explanation:
Objects with more mass have more gravity. Gravity also gets weaker with distance. So, the closer objects are to each other, the stronger their gravitational pull is. Earth's gravity comes from all its mass
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
Answer:
The more hydrogen bonds a molecule can make, the higher the surface tension.
Explanation:
Hydrogen bonds provide higher surface tension to a liquid
More hydrogen molecules - stronger cohesive forces
Answer:
h = 618.64 m
Explanation:
First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:
h₁ = Vit + (1/2)at²
where,
h₁ = height gained during the burning of fuel
Vi = Initial Velocity = 0 m/s
t = time = 7 s
a = acceleration = 8 m/s²
Therefore,
h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²
h₁ = 196 m
Now we use 1st equation of motion to find final speed Vf:
Vf = Vi + at
Vf = 0 m/s + (8 m/s²)(7 s)
Vf = 56 m/s
Now, we calculate height covered in free fall motion. Using 3rd equation of motion:
2ah₂ = Vf² - Vi²
where,
a = - 3.71 m/s²
h₂ = height gained during free fall motion = ?
Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)
Vi = 56 m/s
Therefore,
(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²
h₂ = 422.64 m
So the total height gained will be:
h = h₁ + h₂
h = 196 m + 422.64 m
<u>h = 618.64 m</u>
