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Kitty [74]
3 years ago
8

Describing Forms of Energy

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

Answer:

electrical energy-found in the flow of electrical charges

chemical energy-found in the bonds BTW atoms

thermal energy-found when internal energy is transferred

nuclear energy-found in the nucleus of an atom.

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Why does the air pressure inside the tires of a car increase when the car is driven?
Brrunno [24]
Its because particles of air inside it are moving faster and making more pressure plus there is a pressure from the street.
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xz_007 [3.2K]

Answer: A. All of the answers are correct.

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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t
kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel 40\times 2=80 miles

thus its position vector after two hours is

r_A=-80sin35\hat{i}+80cos35\hat{j}

similarly B travels with 20 mph and in 2 hours

=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}

Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

v_A=-40sin35\hat{i}+40cos35\hat{j}

Velocity of B

v_B=20sin80\hat{i}+20cos80\hat{j}

Velocity of A w.r.t B

v_{AB}=v_A-v_B

v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}

4 0
3 years ago
9. What do people playing pool use to determine their shots?
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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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