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DochEvi [55]
2 years ago
10

An object is dropped from rest and falls through height h. It travels 0.5h in the last 1 second of fall. Find the total time &am

p; height of the fall. (Hint: use two triangles!)
Physics
1 answer:
STALIN [3.7K]2 years ago
4 0

Answer:

3.41 s

114 m

Explanation:

The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.

It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.

0.5*h = 1/2 * a * t1^2

h = a * t1^2

It travels 0.5*h in 1 second.

h = X(t1 + 1) = 1/2 * a * (t1+1)^2

Equating both equations:

a * t1^2 = 1/2 * a * (t1+1)^2

We simplify a and expand the square

t1^2 = 1/2 * (t1^2 + 2*t1 + 1)

t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0

1/2 * t1^2 - t1 - 1/2 = 0

Solving electronically:

t1 = 2.41 s

total time = t1 + 1 = 3.41.

Now

h = a * t1^2

h = 9.81 * 3.41^2 = 114 m

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Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Ne
Levart [38]

Answer:

the number of turns in the primary coil is 13000

Explanation:

Given the data in the question;

V₁ = 13000 V

V₂ = 120 V

N₁ = ?

N₂ = 120 turns

the relation between the voltages and the number of turns in the primary and secondary coils can be expressed as;

V₁/V₂ = N₁/N₂

V₁N₂ = V₂N₁

N₁ = V₁N₂ / V₂

so we substitute

N₁ = (13000 V × 120 turns) / 120 V

N₁ = 1560000 V-turns / 120 V

N₁ = 13000 turns

Therefore, the number of turns in the primary coil is 13000

8 0
2 years ago
What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
PolarNik [594]

Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J

\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

Also, \Delta E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

Also, \Delta E=h\times \nu

So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

8 0
3 years ago
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Taya2010 [7]

Answer:

  • 5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force  + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} ,  mass = density x volume

  • buoyant force =  density x volume x acceleration

        where density is the density of air for the buoyant force

        buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

  • force by helium =  density x volume x acceleration

        force by helium =  0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

  • force of its weight = mass of rubber x acceleration

        force of its weight = 0.0125 x 9.8 = 0.1225 N

  • Tension = buoyant force  + force by helium + force of weight of rubber

         the force  of weight of rubber and of helium act downwards, so they      

          carry a negative sign.

  • Tension = 6.54 - 0.917 - 0.1225 = 5.5 N
8 0
3 years ago
Which of the following is not a property of electromagnetic waves?
lawyer [7]

Answer: It does not include alpha rays

Explanation: It does not include alpha rays

4 0
3 years ago
As an aid in working this problem, consult Concept Simulation 11.1. The volume flow rate in an artery supplying the brain is 3.5
klemol [59]

Answer

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area = π R²

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v = \dfrac{Q}{A}

v = \dfrac{3.5 \times 10^{-6}}{1.06 \times 10^{-4}}

v = 0.033 m/s

b) new velocity of flow

Radius = R' = R/4

 A V = A' V'

 R² V = R'² V'

 R^2 V = (\dfrac{R}{4})^2 V'

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 V' = 16 x 0.033

  V' =0.528 m/s

5 0
2 years ago
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