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Kaylis [27]
3 years ago
12

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of

Physics
1 answer:
Lana71 [14]3 years ago
5 0

Electric field due to a point charge is given as

E = \frac{kq}{r^2}

here we know that

q = 1.6 \times 10^{-19} C

also the distance is given as

r = 5.29 \times 10^{-11} m

now we will have

E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}

so we will have

E = 5.14 \times 10^{11} N/C

so above is the electric field due to proton

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A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0
Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\

If we carry out summation of Torques on the pulley we will have to,

F_2*d-F_1*d = I \alpha

Where,

I = Inertia moment

\alpha =Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

I = \frac{1}{2} mr^2

Replacing this equations we have know that

(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})

F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}

F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1

Or

F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}

Replacing our values we have that

F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}

F_1 = 0.974 N

Therefore the tension in the string between the pulley and the cart is  0.974 N

6 0
4 years ago
The equation for distance is d = st i. If a car has a speed of 22.3 m/s and drives for 8.5 seconds, how far does it go? (1 point
Vladimir79 [104]

distance(d) = speed(s) × time (t)

= 22.3 × 8.5

= 189.55m

6 0
3 years ago
According to science, the fastest speed possible for anything is the speed of light. The speed of light is about 186,000 miles p
Schach [20]
Yes, yes, go on !  The nearest star to us that's outside of our solar system
is 4.2 light years away.  So when we see that star, we're not seeing it as
it is, we're seeing it as it was 4.2 years ago.  And all of the other stars
are much farther away from us than that one is.  We have detected
galaxies that we're not seeing as they are now, we're seeing them as
they were 13 billion years ago !

The moon is about 240,000 miles away from us.  When we look at the
moon, we're not seeing it as it is, we're seeing it as it was 1.3 seconds ago.

The refrigerator in your kitchen is about 12 feet from you.  So when you
look at the refrigerator, you're not seeing it as it is.  You're seeing it as it
was 0.0000000122 second ago.

It takes TIME for light to travel from one place to another.
6 0
3 years ago
Which of the following terms describes the streak of light created when a space object enters Earth's atmosphere?
Burka [1]
Impact crater would be the answer but tell me if I'm roght
5 0
4 years ago
Read 2 more answers
A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average
s344n2d4d5 [400]

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

          a_{average}= 1.5  \ \frac{m}{s^2}

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

           a =   \frac{dv}{dt}

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

          v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make  the derivative.

           a = α + 2β t

Let's substitute

            a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.  

          a_{average} = \frac{\Delta v}{\Delta t }

Let's find the velocity at the indicated time.

For t = 5 s

         v₅ = 3 + 0.1 5²

         v₅ = 5.5 m / s

For t = 10 s

          v₁₀ = 3 + 0.1 10²

          v₁₀ = 13 m / s

Let's calculate the average acceleration.

           a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\

           a_{average}= 1.5 \  m/s^2

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

           a_{average} =  1.5 m / s²

Learn more here: brainly.com/question/20057878

7 0
3 years ago
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