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Nikitich [7]
3 years ago
7

An object is traveling with a constant velocity of 5 m/s. How far will it have gone after 7 s?

Physics
1 answer:
jeka943 years ago
8 0

Answer:

35 miles

Explanation:

5 times 7

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What scientific theory was the subject of the scopes trial?
ICE Princess25 [194]
Scopes, who has substituted for the regular biology teacher was charged on May the 5th, 1925 with teaching evolution from a chapter in George William Hunters textbook. Civic Biology: Presented in problems which described the theory of evolution... Hope this helps!
7 0
3 years ago
The graph shows a car's velocity over time. During which time period does the car have the greatest acceleration?
Oksana_A [137]

Answer:

Between 0 and 1 seconds (B)

Explanation:

The velocity of the car over time is represented by the line graphed here

the steeper the line, the greater change in velocity that occurred in a given time frame.

The steepest portion of the line is between 0-1 seconds, which means that the greatest rate of change occurred between 0-1 seconds.

(acceleration is the rate of change)

5 0
2 years ago
Read 2 more answers
Andrea asked her brother to take a 4 ft floating raft out of the water near the wave-swept shore. Using this raft as a measuring
lutik1710 [3]

Answer:

176ft/s

Explanation:

Speed of a wave is a function of the frequency and wavelength of the wave. It is expressed mathematically as:

V = fλ where:

V is the speed of the wave

f is the frequency

λ is the wavelength

Given f = 16Hz, λ = 11ft

V = 16×11

V = 176ft/s

7 0
4 years ago
Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnit
Snowcat [4.5K]

Answer:

894 electrons

Explanation:

The electrostatic force between the two charges is given by:

F=\frac{k q_1 q_2}{r^2}

where we have

F=4.57\cdot 10^{-21} N is the force

k is the Coulomb's constant

q1 = q2 =q is the magnitude of the charge on each sphere

r = 20.0 cm = 0.20 m is the distance between the two spheres

Substituting and solving for q, we find the charge on each sphere:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(4.57\cdot 10^{-21} N)(0.20 m)^2}{9\cdot 10^9 Nm^2C^{-2}}}=1.43\cdot 10^{-16} C

And since each electron has a charge of

e=1.6\cdot 10^{-19}C

the net charge on each sphere will be given by

q=Ne

where N is the number of excess electrons; solving for N,

N=\frac{q}{e}=\frac{1.43\cdot 10^{-16}C}{1.6\cdot 10^{-19}C}=894

7 0
4 years ago
Part A: Determine the wavelength of photons that can be emitted
torisob [31]

Answer:

A  λ = 97.23 nm

, B)   λ = 486.2 nm

, C)  λ = 53326 nm

Explanation:

With that problem let's use the Bohr model equation for the hydrogen atom

          E_{n} = -k e² /2a₀  1/n²

For a transition between two states we have

          E_{nf} -  E_{no} = -k e² /2a₀ (1/  n_{f}² - 1 / n₀²)

Now this energy is given by the Planck equation

         E = h f

And the speed of light is

         c = λ f

Let's replace

      h c / λ = - k e² /2a₀ (1 / n_{f}² - 1 / no₀²)

      1 /  λ = - k e² /2a₀ hc (1 / n_{f}² -1 / n₀²)

Where the constants are the Rydberg constant R_{H} = 1.097 10⁷ m⁻¹

        1 /  λ = R_{H} (1 / n₀² - 1 / nf²)

Now we can substitute the given values

Part A

 Initial state n₀ = 1 to the final state n_{f} = 4

        1 /  λ = 1.097 10⁷ (1/1 - 1/4²)

         1 /  λ = 1.0284 10⁷ m⁻¹

          λ = 9.723 10⁻⁸ m

We reduce to nm

         λ = 9.723 10⁻⁸ m (10⁹ nm / 1m)

         λ = 97.23 nm

Part B

Initial state n₀ = 2 final staten_{f} = 4

       1 /  λ = 1.097 10⁷ (1/2² - 1/4²)

       1 /  λ = 0.2056 10⁻⁷ m

        λ = 486.2 nm

Part C

Initial state n₀ = 3

      1 /  λ = 1,097 10⁷ (1/3² - 1/4²)

       1 /  λ = 5.3326 10⁵ m⁻¹

        λ = 5.3326 10-5 m

        λ = 53326 nm

5 0
3 years ago
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