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tangare [24]
3 years ago
10

Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m

ass of 0.5 kilograms, falls off the edge of the scaffolding. Calculate the bucket’s kinetic and potential energy when it is 4 meters above the ground. Also calculate its velocity at this point.

Physics
2 answers:
saul85 [17]3 years ago
7 0

Answer:

The bucket's velocity is v = 12.522 m/s

Explanation:

Given

Height above the ground, H = 12 m

Mass of bucket, m = 0.5 kg

Height of bucket, h = 4m

Taking acceleration of gravity, g as 9.8m/s²

First, we calculate the potential energy when bucket is on scaffolding (just about to fall) its total energy is 100%

P.E = mgH

P.E = 0.5 * 9.8 * 12

P.E = 58.8 J

At this point, the object is at rest, so the kinetic energy is 0 J

When the object is at 4 m above the ground, the bucket's total mechanical energy is as follows using conservation of energy

M.E = mgh + \frac{1}{2} mv^{2}

Where Total mechanical energy (M.E) = P.E (calculated above) = 58.8 J

By substitution, we have

58.8 = 0.5 * 9.8 * 4 + \frac{1}{2} * 0.5 * v^{2}

58.8 = 19.6 + \frac{1}{2} * 0.5 * v^{2}

58.8 - 19.6 = \frac{1}{2} * 0.5 * v^{2}

39.2 = \frac{1}{2} * 0.5 * v^{2}

39.2 = \frac{1}{2} * \frac{1}{2} * v^{2}

39.2 * 2 * 2=  v^{2}

156.8=  v^{2} --- Take Square roots

\sqrt{156.8} =  \sqrt{v^{2}}

v = \sqrt{156.8}

v = 12.522 m/s ---- Approximated

Hence, the bucket's velocity is v = 12.522 m/s

Sever21 [200]3 years ago
3 0

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:  

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12        ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4        ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

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E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

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C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

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∠Y = tan⁻¹(0.4/0.3) = 53.13°

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Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

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Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

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3 years ago
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Answer:

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Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

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D is the diffusivity coefficient

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In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

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Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

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\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

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J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

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