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astra-53 [7]
3 years ago
6

Two students, Jordan and Taylor, are talking about light. Jordan says that light behaves as a stream of particles. Taylor declar

es that light behaves as a wave. Which student is correct?
Physics
1 answer:
FrozenT [24]3 years ago
3 0

Answer:

"Both students are correct. When light is traveling through small openings, solids, liquids, gases, or a vacuum, it behaves like a wave. When it interacts with matter, it acts like a stream of particles."

Explanation:

just got it right

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What force causes a 1-kg mass to accelerate at a rate of 1 meter per second each second?
Zigmanuir [339]

Answer:

F=1N

Explanation:

Conceptual analysis

To solve this problem we apply Newton's second law:

The acceleration of an object is proportional to the force F acting on it and inversely proportional to its mass m.

a = F / m

Where,

F = m * a Formula (1)

F: Force in Newtons (N)

m: mass in kg

a: acceleration in m/(s^2)

a = v / t  Formula (2)

v: speed in m/s

t: time in seconds (s)

Known information

We know the following data:

m = 1kg

v = 1 m/s

t = 1s

Development of the problem:

In the Formula (2): a = \frac{\frac{1m}{s}}{1s} = 1 \frac{m}{s^2}

In the Formula (1): F=1kg* 1 \frac{m}{s^2}=1N

8 0
3 years ago
1. When two atoms of the same nonmetal react, they often form a(an)
kkurt [141]

When two atoms of the same nonmetal react,they form what we know today as a diatomic molecule.

Please mark brainliest. :)

7 0
3 years ago
What will happen to the force felt between two charged objects if the distance between them is 1/3rd of the original distance
kenny6666 [7]

Answer:

New force = 9(initial force)

Explanation:

The force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

d is the original distance

Let d' is the new distance such that, r' = r/3

New force,

F'=\dfrac{kq_1q_2}{r'^2}\\\\F'=\dfrac{kq_1q_2}{(\dfrac{r}{3})^2}\\\\F'=9\times \dfrac{kq_1q_2}{r^2}\\\\F'=9F

So, the new force becomes 9 times the initial force.

4 0
3 years ago
Assuming this is a distance time graph( ignore the speed time title) assume metres on vertical scale. describe in as much detail
riadik2000 [5.3K]

Answer:

The journey started from point A, with a speed of  1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance travelled during the journey from the start point A to the final point B is 40 m

Explanation:

From the start point A to point B, we have;

The speed from A to B = 10 m/(10 s) = 1 m/s

The distance traveled from A to B = 10 m

The time it takes to move from A to B = 10 seconds

From the point B to point C, we have;

The distance traveled from B to C = 0 m, (stationary)

The time it remains at point B distance from the start point = 10 seconds

The speed between point B to C = 0 m/(10 s) = 0 m/s

From the point C to point D, we have;

The distance traveled from C to D = 10 m

The time it takes to move from C to D = 5 seconds

The speed between point C and D = 10 m/(5 s) = 2 m/s

From the point D to point E, we have;

The distance traveled from D to E = 0 m, (stationary)

The time it remains at point D distance from the start point = 10 seconds

The speed between point D to E = 0 m/(10 s) = 0 m/s

From the point E to point F, we have;

The distance traveled from E to F = 20 m (return journey starts at point E)

The time it takes to move from E to F = 5 seconds

The speed between point E to F = 20 m/(5 s) = 4 m/s (Return journey)

Therefore, the journey started from point A, with a speed of  1 m/s for 10 seconds after which the it became stationary at 10 meters from the start point, for 10 seconds. The journey continued at a higher speed of 2 m/s for 5 seconds and halted (became stationary) again at 20 meters which was the maximum distance reached from the start point

The return journey to the start point started at the 20 meter mark and lasted for 5 seconds, at a speed of 4 m/s

The total distance moved, 'd', to and from the start point with reference to the graph is given as follows;

d = (From A to B) 10 m + (From B to C) 0 m + (From C to D) 10 m + (From D to E) 0 m + (From E to F) 20 m = 40 m

The total distance travelled in the journey is 40 m

The total displacement, \underset{d}{\rightarrow} = 10 m + 10 m - 20 m = 0 m

7 0
3 years ago
You slide a slab of dielectric between the plates of a parallel-plate capacitor. As you do this, the potential difference betwee
LUCKY_DIMON [66]

Answer:

In this scenario adding the dielectric material in between the plates will have no effect on the capacitance of the plates since the voltage remains unchanged

Explanation:

Normally Introducing a dielectric into a capacitor decreases the electric field, which decreases the voltage, which increases the capacitance.

A capacitor with a dielectric stores the same charge as one without a dielectric, but at a lower voltage.

Voltage and capacitance are inversely proportional when charge is constant.

Now in this case the voltage remains the same hence the charges remain the same also because voltage is inversely proportional to capacitance

3 0
4 years ago
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