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kipiarov [429]
3 years ago
11

How could two waves on a rope interfere so that the rope did not move at all.

Physics
1 answer:
Minchanka [31]3 years ago
8 0
2 waves will overlap  

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Which scenario is an example of the transfer of thermal energy by radiation?
Mumz [18]

Answer:the answer is B

Explanation: just took the quiz

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2.
horrorfan [7]

Answer:

B. 6HgO → 6Hg + 3O_{2}

Explanation:

A decomposition reaction is a reaction in which a single reactant is broken down into 2 or more products.

6 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

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        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
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Janet ran 30 meters in 4 seconds,what is her speed.
lubasha [3.4K]

Answer:

7.5 meters per second

Explanation:

30 / 4 = 7.5

7.5 * 1 = 7.5

7.5 * 2 = 15

7.5 * 3 = 22.5

7.5 * 4 = 30

8 0
3 years ago
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The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.
irina [24]

Part a)

At t = 0  the position of the object is given as

x = 0

At t = 2

x = 2 sin(\pi/2) = 2cm

so displacement of the object is given as

d = 2 - 0 = 2cm

so average speed is given as

v_{avg} = \frac{2}{2} = 1 cm/s

Part b)

instantaneous speed is given by

v = \frac{dy}{dt}

v = 2cos(\pi t/4 ) * \frac{\pi}{4}

now at t= 0

v = \frac{\pi}{2} cm/s

at t = 1

v = 2 cos(\pi/4) * \frac{\pi}{4}

v = \frac{\pi}{2\sqrt2}

at t = 2

v = 0

Part c)

Average acceleration is given as

a_{avg} = \frac{v_f - v_i}{t}

a_{avg} = \frac{0 - \frac{\pi}{2}}{2}

a = -\frac{\pi}{4} cm/s^2

Part d)

Now for instantaneous acceleration

As we know that

a =- \omega^2 y

at t = 0

a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2

at t = 1

y = \sqrt2 cm

now we have

a = -\frac{\pi^2}{16}*\sqrt2

At t = 2 we have

y = 2 cm

a = -\frac{\pi^2}{16}*2

a = -\frac{\pi^2}{8}

<em>so above is the instantaneous accelerations</em>

7 0
3 years ago
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