Which changes in energy form are illustrated in the diagram.

Physics

1 answer:

Grace [21]2 years ago

6 0

It would be B since it starts with the solar energy which is converted to electricity with the solar panels, which then creates mechanical energy for the fans blades to move and sound for the radio.

Hope that helps :)

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If the hiker starts climbing at an elevation of 350 ft, what will their change in gravitational potential energy be, in joules,

Kitty [74]

Answer:

352,088.37888Joules

Explanation:

Complete question;

A hiker of mass 53 kg is going to climb a mountain with elevation 2,574 ft.

A) If the hiker starts climbing at an elevation of 350 ft., what will their change in gravitational potential energy be, in joules, once they reach the top? (Assume the zero of gravitational potential is at sea level)

Chane in potential energy is expressed as;

ΔGPH = mgΔH

m is the mass of the hiker

g is the acceleration due to gravity;

ΔH is the change in height

Given

m = 53kg

g = 9.8m/s²

ΔH = 2574-350 = 2224ft

since 1ft = 0.3048m

2224ft = (2224*0.3048)m = 677.8752m

Required

Gravitational potential energy

Substitute the values into the formula;

ΔGPH = mgΔH

ΔGPH = 53(9.8)(677.8752)

ΔGPH = 352,088.37888Joules

Hence the gravitational potential energy is 352,088.37888Joules

7 0

2 years ago

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$