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jasenka [17]
3 years ago
14

How fast must a 2500-kg elephant move to have the same kinetic energy as a 67.0-kg sprinter running at 15.0 m/s

Physics
1 answer:
abruzzese [7]3 years ago
6 0

Answer:

2.45 m/s

Explanation:

kinetic energy = 1/2 * m * v^2

then, 0.5 * 2500 * x^2 = 0.5 * 67 * 15^2

by solving for x, X = 2.45 m/s

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What is this question formula a=V²-U² /2S?​
erik [133]

Answer:

sorry I don't really know about that question.

Explanation:

7 0
2 years ago
A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com
blagie [28]

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

8 0
3 years ago
A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente
nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is I_3 =41.31 W/m^2

Explanation:

 According to Malu's law the intensity of a polarized light having an initial intensity I_0 is mathematically represented as

               I = I_0cos^2 \theta

Now  considering the polarizer(The polarizing disk) the equation above becomes

          I = I_0 (cos^2 \theta)^n

Where n is the number of polarizers

       Substituting  60.0W/m^2 for the initial intensity 3 for the n and 20° for the angle of rotation

           I_3 = 60 (cos^220)^3

               =41.31 W/m^2

             

     

                         

6 0
3 years ago
To balance a chemical equation we use______. <br> -coefficients <br> -subscripts
Vitek1552 [10]

Answer:

Coefficients

Explanation:

  • Coefficients are used to balance chemical equations.
  • According to the law of conservation of mass, the mass of the reactants in a chemical equation should be equivalent to the mass of the products.
  • The conservation of mass is achieved in chemical equations is achieved through balancing chemical equations.
  • <em><u>Balancing chemical equations is a try and error of putting appropriate coefficients to the reactants or products to ensure that the number of atoms of each element are equal on the side of the reactants and that of products. </u></em>
3 0
3 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

4 0
3 years ago
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