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prohojiy [21]
3 years ago
5

Can somebody please work these questions out for me?.

Physics
1 answer:
ZanzabumX [31]3 years ago
5 0

Answer:

Please find the calculated values entered into the following table;

\left\begin{array}{ccccc}Object & Mass \ (kg)& Weight  \ (N) & Height \ (m)&G.P.E.\\A&5&50&2&100\\B&2&20&6&120\\C&8&80&5&400\\D&20&200&0.6&120\\E&5,000&50,000&2&100,000\\F&0.2&2&10&20\\G&67&670&44&29,480\end{array}

Explanation:

The given parameters are;

The acceleration due to gravity, g ≈ 10 N/kg

The Gravitational Potential Energy, G.P.E. = m × g × h

The weight of the object, W = m × g

∴ G.P.E. = W × h

The table is filled using the above formula for calculations as follows;

For object A, we have;

Weight = 5 kg × 10 N/kg = 50 N

G.P.E. = 50 N × 2 m = 100 J

For object B, we have;

Weight = 2 kg × 10 N/kg = 20 N

G.P.E. = 20 N × 6 m = 120 J

For object C, we have;

Weight = 8 kg × 10 N/kg = 80N

G.P.E. = 80 N × 5 m = 400 J

For object D, we have;

Weight = 20 kg × 10 N/kg = 200 N

G.P.E. = 200 N × 0.6 m = 120 J

For object E, we have;

Weight = 5,000 kg × 10 N/kg = 50,000 N

G.P.E. = 50,000N × 2 m = 100,000 J

For object F, we have;

Weight = 0.2 kg × 10 N/kg = 2 N

G.P.E. = 2 N × 10 m = 20 J

For object G, we have;

Weight = 67 kg × 10 N/kg = 670 N

G.P.E. = 670 N × 44 m = 29,480 J

Therefore, we have the table filled as follows;

\left\begin{array}{ccccc}Object & Mass \ (kg)& Weight  \ (N) & Height \ (m)&G.P.E.\\A&5&50&2&100\\B&2&20&6&120\\C&8&80&5&400\\D&20&200&0.6&120\\E&5,000&50,000&2&100,000\\F&0.2&2&10&20\\G&67&670&44&29,480\end{array}

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When a certain gas under a pressure of 5.00 × 106 Pa at 25.0°C is allowed to expand to 3.00 times its original volume, its final
12345 [234]

Answer:

191.316 K or -81.684 °C

Explanation:

From general gas law,

P₁V₁/T₁ = P₂V₂/T₂ ................ Equation 1

Where P₁ = Initial pressure, V₁ = Initial volume, T₁ = Initial temperature, P₂ = Final pressure, V₂ = Final volume, T₂ = Final Temperature.

Make T₂ the subject of the equation.

T₂ = P₂V₂T₁/P₁ V₁ ............... Equation 2

Given: P₁ = 5.00×10⁶ Pa, T₁ = 25.0°C = 298 K, P₂ = 1.07×10⁶.

Let: V₁ = y cm³, V₂ = 3y cm³

Substitute into equation 2,

T₂ = (1.07×10⁶×298×3y)/(5.00×10⁶×y)

T₂ =191.316 K.

Hence the final temperature = 191.316 K or -81.684 °C

4 0
3 years ago
A single-slit diffraction pattern is formed on a distant screen. If the width of the single slit through which light passes is r
anygoal [31]

Answer:

Explanation:

General guidance

Concepts and reason

The concept used to solve this problem is slit width condition for maximum diffraction in case of single slit diffraction experiment.

Initially, use the condition for diffraction maximum in the case of single slit diffraction to find the inapplicable given options.

Finally, use the condition for diffraction maximum in the case of single slit diffraction to find the applicable given options.

Fundamentals

The condition for diffraction maximum in the case of single slit diffraction is as follows:

sin Θ=λ/α

Here, the angle situated in the first dark fringe on each side of the central bright fringe isΘ , slit width is α, and the wavelength is λ .

The incorrect options are as follows:

• The central bright fringe remains in same size.

The width of the central bright fringe is inversely proportional to the slit width. Therefore, the central fringe cannot remain in the same size. Hence, it is incorrect.

• The effect cannot be determined unless the distance between the slit and screen is known.

Without knowing the distance between the slit and screen, the effect can be experienced. Therefore, this option is incorrect.

• The central bright fringe becomes narrower.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becoming narrower is incorrect.The central bright fringe width is directly proportional to the slit width.

If the width of the slit increases, then the central bright fringe width also increases.

Due to the inverse proportion of central bright fringe width and slit width, the central bright fringe becomes wider when the width of the single slit is reduced.

The condition for diffraction maximum is as follows:

sin  Θ=λ/α

The slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

Therefore,

• The central bright fringe becomes wider is correct.

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.  

Slit width is inversely proportional to the angle of the first dark fringe on either side of the central bright fringe.

The central bright fringe width is directly proportional to the angle of the first dark fringe on either side of the central bright fringe.

If the central bright fringe becomes wider, then the angle of the first dark fringe on either side of the central bright fringe will be larger.

Answer

The applicable option when the width of the slit reduces is the central bright fringe becoming wider.

5 0
4 years ago
A student watches as a trash can lid moves across a yard. Which of these could have caused the trash can lid to move?
omeli [17]

Answer:

wind energy

Explanation:

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In the context, trash can lid is being blown away by the wind energy and it is seen moving across the yard. The energy of the wind forces the lid of the trash can to move from one place to another against friction. Thus, wind energy caused the trash can lid to move across the yard as seen by a student.

8 0
3 years ago
A record of travel along a straight path is as follows:
nignag [31]

Answer:

a) Total displacement  = 3986.54 m

b) Average speeds

      Leg 1 ->  11.22 m/s

      Leg 2 ->  22.44 m/s

      Leg 3 ->  11.20 m/s

      Complete trip ->  21.63 m/s

Explanation:

a) Leg 1:

Initial velocity, u =  0 m/s

Acceleration , a = 2.04 m/s²

Time, t = 11 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    s = 0 x 11 + 0.5 x 2.04 x 11²

    s = 123.42 m

Leg 2:

We have equation of motion v = u + at

Initial velocity, u =  0 m/s

Acceleration , a = 2.04 m/s²

Time, t = 11 s

Substituting

   v = 0 + 2.04 x 11 = 22.44 m/s

We have equation of motion s= ut + 0.5 at²

Initial velocity, u =  22.44 m/s

Acceleration , a = 0 m/s²

Time, t = 2.85 min = 171 s

Substituting

   s= ut + 0.5 at²

    s = 22.44 x 171 + 0.5 x 0 x 171²

    s = 3837.24 m

a) Leg 3:

Initial velocity, u =  22.44 m/s

Acceleration , a = -9.73 m/s²

Time, t = 2.31 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    s = 22.44 x 2.31 + 0.5 x -9.73 x 2.31²

    s = 25.88 m

Total displacement = 123.42 + 3837.24 + 25.88 = 3986.54 m

Average speed is the ratio of distance to time.

b) Leg 1:

        v_{avg}=\frac{123.42}{11}=11.22m/s

 Leg 2:

        v_{avg}=\frac{3837.24}{171}=22.44m/s

Leg 3:

        v_{avg}=\frac{25.88}{2.31}=11.20m/s

Complete trip:

        v_{avg}=\frac{3986.54}{11+171+2.31}=21.63m/s

                           

5 0
3 years ago
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ZanzabumX [31]

Answer:

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