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vodomira [7]
3 years ago
7

Calculating acceleration worksheet

Physics
1 answer:
MrRissso [65]3 years ago
8 0
Yes bc math, numbers and more
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9. A 12 v battery is connected to four 5 ohm light bulbs. What is the equivalent
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20 ohms 5 ohms
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2 years ago
The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an
blsea [12.9K]

Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

        F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

      F = G Me /Re²  m

We call gravity acceleration a

       g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

            R = Re + h

Therefore  the attractive force is

      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

       F = g₀   m  [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

     m g =g₀ m  [1- 2 h / Re + 3 (h / Re)²]

       g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

3 0
3 years ago
Which part of the water cycle would follow step D in the diagram shown?
Dmitry [639]

Condensation.


iGreen, more like iNOTGreen amirite?

5 0
3 years ago
Read 2 more answers
What is the definition of Mutualism
noname [10]
Mutualism is a long-term relationship where two organisms interact in such a way that both of them benefits from that relationship.

For example, there is a relationship between a bird called "oxpecker", and a rhino. While the bird eats the harmful bugs (eg. tick) on the rhino's skin and relieves its hunger; the rhino gets rid of the bugs that harm it.
5 0
3 years ago
Read 2 more answers
A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
3 years ago
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