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Nana76 [90]
3 years ago
5

The magnitude of the net electrical force acting on the +6 µC charge, rounded to the tenths place, is N.

Physics
2 answers:
Lyrx [107]3 years ago
8 0
Electrical force is the repulsive or attractive force of attraction between any two charged bodies. it is possible to calculate electric force on anything that runs on batteries or uses a plug. Its net effect can be explained using newtons law of motion like any force.
 From the coulombs law; F = k (q1║q2)/ r²
                       Where k = 9 ×10∧9 N.m².C∧-2, r12= 0.285m, q2= +6μC and q1 = 4.86 nC
substituting values in the equation
 the net electrical force acting on the +6ц C charge is 5.4 N.
blondinia [14]3 years ago
4 0

Answer:

5.4 N

Explanation:

Got it right on edge

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Mary cycled at an average speed of 8 km/h. How far has she traveled if she rides for 4 hours?
Sati [7]

Answer:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

Explanation:

Which sentence from the passage shows that the function of the river depicted here has carried through to modern times?

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3 years ago
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A proton is confined within an atomic nucleus of diameter 3.60 fm. part a estimate the smallest range of speeds you might find f
Cerrena [4.2K]
The answer for this problem would be:
Assuming non-relativistic momentum, then you have: 
ΔxΔp = mΔxΔv = h / (4) 
Δv = h / (4πmΔx) 
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 --> 
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s 
That's about 1% of the speed of light, the assumption that it's non-relativistic.
3 0
3 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

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7 0
3 years ago
A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask
Charra [1.4K]

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

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