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Slav-nsk [51]
2 years ago
6

Please help me I will give a brainless

Physics
1 answer:
gulaghasi [49]2 years ago
4 0

Answer:

By opening the door, Elijah is using the chemical energy, which he obtained from eating food to do work. Thus, he is converting chemical energy to mechanical energy.

Stretching the spring by opening the screen door convert the potential energy which is stored in the spring to kinetic energy.

Overall there is conversion of chemical energy to potential energy.

By letting the door go, Elijah is using chemical energy. By snapping shut, the door is converting potential energy to kinetic energy. The overall energy change is conversion of chemical energy to kinetic energy.

Explanation:

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The name of the Sl unit for force is the ___?<br><br> A. Joule<br> B. Newton<br> C.Watt
tiny-mole [99]

Answer:

the name of the SI unit for force is the newton

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What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k
Molodets [167]

Answer:

\dfrac{I_1}{I_2}=12.25

Explanation:

r_1 = 14 km

r_2 = 49 km

Intensity of a wave is inversely proportional to distance

I\propto \dfrac{1}{r^2}

So,

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25

The ratio of the intensities is \dfrac{I_1}{I_2}=12.25

6 0
3 years ago
a ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base. What was the ball´s initia
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Determine the energy, in electron volts, of an emitted photon when an electron transition from n=3 to n=2 occurs
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7 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
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