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Drupady [299]
3 years ago
10

3. What can you infer about an ion that has a positive charge?

Physics
1 answer:
IRINA_888 [86]3 years ago
4 0

Answer: This ion is derived from atom who lost one electron.

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Calculate using coulombs: an object has 6 protons and 8 electrons. Calculate the magnitude of the charge of the objects. Thank y
SVEN [57.7K]

Answer:

-3.2 ×1.6 ×10^-19C

Explanation:

Number of proton = 6

Number of electron = 8

To determine the quantity of the charge, we would first find the difference between the number of

protons and electrons. Then multiply the difference by charge 1.6 x 10 -19 C to determine the charge on the object.

Difference between proton and electron is elementary charge

Elementary charge = proton - electron

= 6 - 8

Elementary charge = -2e

Recall, charge (q) = 1.6 ×10^-19C

1e = 1.6 ×10^-19C

Magnitude of the charge of object = -2× (1.6 ×10^-19C)

Magnitude of the charge of object = -3.2 ×1.6 ×10^-19C

8 0
3 years ago
Answer this question, please​
pashok25 [27]

Answer:

the correct answer is the 60

+ 20 + 60 \gamma  \beta

3 0
3 years ago
Daisy made the graph shown to display the data recorded during an experiment. Daisy is studying the effect of soil on the growth
andreev551 [17]

im in flvs too if thats what this is but anyway im doing it right now and i believe it is sunlight was not kept constant


3 0
3 years ago
Read 2 more answers
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
4 years ago
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
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