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lozanna [386]
4 years ago
8

What keeps the electrons from leaving an atom in the rutherford model of the atom?

Physics
1 answer:
alexdok [17]4 years ago
7 0

A. strong electrostatic attraction to the nucleus


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A car is a compound machine true or false
Viktor [21]
True is the answer. I believe
6 0
3 years ago
f 720-nm and 620-nm light passes through two slits 0.68 mm apart, how far apart are the second-order fringes for these two wavel
valkas [14]

Answer:

0.0003 m = 0.3 mm

Explanation:

For constructive interference in the Young's experiment.

The position of the mth fringe from the central fringe is given by

y = L(mλ/d)

λ = wavelength = 720 nm = 720 × 10⁻⁹ m

L = distance between slits and screen respectively = 1.0 m

d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m

m = 2

y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm

For the 620 nm light,

y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm

Distance apart = 2.12 - 1.82 = 0.3 mm = 0.0003 m

8 0
3 years ago
Why is carrying for the earth necessary? explain​
monitta

Answer: We will survive

Explanation: Caring for the earth is something that we need to do to live or survive on earth because if we don't we may not survive.

Hope this helps! :)

6 0
3 years ago
Read 2 more answers
In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (
FromTheMoon [43]

Answer:

the acceleration a^{\to} = (0.0159 \ \ m/s^2 )i

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So;  the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration  a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}

a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}

a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}

a =0.0159 \ m/s^2

Thus;

Assume the car moves in the +x direction;

the acceleration a^{\to} = (0.0159 \ \ m/s^2 )i

7 0
3 years ago
An object 82 cm high forms a virtual image 4.1 cm high located 4.6 cm behind a mirror. Find the object distance.
ioda

Answer:

The object distance is 92 cm.  

Explanation:

let v be the image distance and h be the height of the image, let u the be the object distance and H be the height of the object.

then, the magification of the mirror is given by:

m = -v/u and m = h/H

so, -v/u = h/H

         u = -v×H/h

            = -(-4.6)×(82)/(4.1)

            = 92 cm

Therefore, the object distance is 92 cm.

8 0
3 years ago
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