The speed a spherical raindrop would achieve falling from 3950 m in the absence of air drag is calculated by firstly finding the time it takes for the raindrop to fall this distance using equations of motion, and then using this time in the equation for final velocity. The calculated speed is approximately 2785.30 m/s.
To calculate the speed a spherical raindrop would achieve falling from 3950 m in the absence of air drag, we must recall the equations of motion. The relevant equation here is Final velocity (v) = Initial velocity (u) + Acceleration (gravity, g) * time (t). However, since initial velocity (u) is 0 (when the drop starts falling, it's stationary), the equation simplifies to Final velocity (v) = g * t.
In free fall, a body accelerates under gravity (approximated as 9.81 m/s^2). In terms of time, difficulties arise because we don't know exactly when the raindrop will hit the ground. We can, however, calculate the time it would take for the raindrop to fall 3950 m by rearranging the equation distance (s) = ut + 0.5 * g * t^2 to solve for time. Removing (u), for the reasons explained earlier, we have the equation s = 0.5 * g * t^2. Solving this for time gives t = sqrt(s / (0.5 * g)). Substituting the given fall distance for s we get t = sqrt(3950 / (0.5 * 9.81)) or approximately 284.10 seconds.
Finally, we use this calculated time in our simplified velocity equation which gives v = g * t or 9.81 * 284.10, which equals approximately 2785.30 m/s.
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2X + 2Z = 10Y
Z equals 10, when substitute X = 5 and Y = 3 in 2X + 2Z = 10Y.
Arithmetic operations can also be specified by the subtract, divide, and multiply built-in functions.
* Multiplication operation: Multiplies values on either side of the operator
For example 4*2 = 8
/ Division operation: Divides left-hand operand by right-hand operand
For example 4/2 = 2
Given that equation as:
⇒ 2X + 2Z = 10Y
If X = 5 and Y = 3,
Substitute the values of X = 5 and Y = 3,
⇒ 2(5) + 2 (Z) = 10(3)
Apply the multiplication operation
⇒ 10 + 2Z= 30
⇒ 2Z = 20
⇒ Z = 20/2
Apply the division operation
⇒ Z = 10
Hence, Z equals 10, when Substitute X = 5 and Y = 3 in 2X + 2Z = 10Y
Learn more about Arithmetic operations here:
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(B) Enter an expression for the angular speed w of the system immediately after the collision, in terms of m, V, D, 0,
(C) Calculate the rotational kinetic energy, in joules, of the system after the collision
We calculate the total moment of inertia of the rod-ball system after the collision by adding the moment of inertia of the rod and the added contribution from the putty ball. With this, we find the post-collision angular speed using Conservation of Angular Momentum. The rotational kinetic energy is then determined from this angular speed.
To solve this problem, we first need to calculate the moment of inertia of the combined system of the rod and the putty. The moment of inertia of an object is given by its mass times the square of its distance from the axis of rotation. That gives us I = 1/3 ML2 + m(D + L/2)2
Next we use Conservation of Angular Momentum to find the post-collision angular speed (ω). The initial momentum (mVD) is equal to the final moment of inertia times the final angular speed, so ω = mVD / I.
Finally, we calculate the rotational kinetic energy, which is given by ½ I ω2.
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b. A planet's orbit covers equal distances in equal amounts of time; the speed of a planet's orbit depends on its distance from the Sun; the bigger the planet, the slower it moves.
c. Planets orbit in elliptical patterns; a planet's orbit covers equal areas in equal amounts of time; planets' orbits are shorter or longer depending on their distance from the Sun.
The correct answer to the question is : C) Planets orbit in elliptical patterns; a planet's orbit covers equal areas in equal amounts of time; planets' orbits are shorter or longer depending on their distance from the Sun.
EXPLANATION:
Before coming into any conclusion, first we have to understand three laws of Kepler on planetary motion.
First law: Every planets moves around the sun in elliptical orbits with the sun situated at one of their foci.
Second law: Planets sweep out equal areas in equal interval of time.
Third law: The square of time period of rotation of every planet is directly proportional to the cube of semi major axis.
Hence, from above we see that the option three corresponds to the Kepler's laws partially.
Hence, the correct statement is option three.
b. It is divided into a pair of fives.
c. It has a stressed syllable followed by an unstressed syllable.
d. It is present in both Italian sonnets and Shakespearean sonnets.