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Vladimir [108]
3 years ago
15

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Physics
1 answer:
Virty [35]3 years ago
7 0

Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
  • g= 9.8 m/s²
  • h= 20 m

Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

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Rom4ik [11]

Answer:

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Explanation:

From the question,

The potential energy of the bird = heat gained by the water in the fish tank.

mgh = cm'(Δt)................... Equation 1

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make Δt the subject of the equation

Δt = mgh/cm'............... Equation 2

Given: m = 1 kg, h = 40 m, m' = 50.5 kg

constant: g = 9.8 m/s², c = 4200 J/kg.K

Substitute into equation 2

Δt = 1(40)(9.8)/(50.5×4200)

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

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     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

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   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

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Anna35 [415]

Answer:

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Explanation:

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2 years ago
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