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Andreyy89
3 years ago
11

Two iron rods are placed horizontally, touching end-to-end. One end of one rod is heated. After some time, the other rod also fe

els warm to touch. How did the heat travel from the heated rod to the one that was not heated?
A) The air over the heated rod traveled to the other rod and made it warm.
B) Part of the heat was converted to radiation and traveled to the cooler rod.
C) Electrons within the rods started moving, distributing heat consistently in both rods.
D) The molecular movement in the first rod transferred energy to the molecules in the second.
Physics
2 answers:
algol133 years ago
6 0
The answer is D) <span>The molecular movement in the first rod transferred energy to the molecules in the second.</span>
saw5 [17]3 years ago
3 0

I can confirm that the answer is D

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g&#10;            y- 0 = 10.0²/2 9.8&#10;            y - 0 = 5.10 m&#10;            &#10;The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system&#10;             y₂ = 5.1 + 44&#10;             y₂ = 49.1 m&#10;Let's use the other equation to find the time&#10;              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Alex Ar [27]

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

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        a = F / m

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        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

3 0
3 years ago
Read 2 more answers
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