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3 years ago

8

What is a electron orbit?

2 answers:

skelet666 [1.2K]3 years ago

7 0

Electrons occupy shells around the nucleus of an atom. Electrons move in every direction, but they are limited to their own area, or the orbit that the electron follows, which is what we call shells. Within each shell, there are sub-shells. ... Orbitals are regions within an atom that the electron will most likely occupy.

Advocard [28]3 years ago

5 0

the path of an electron around the nucleus of an atom

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A 2,000 kg car is parked at the top of a 30 m high hill. what is its potential energy?

Phoenix [80]

The PE for this question will be 588,000 because we take the mass (2,000 kg), multiply it by 9.8 which is Gravitational Acceleration and then multiply that by the height (30 meters)

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3 years ago

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two horses pull against a rope with forces of 100 newtons in opposite directions. this is an example of

statuscvo [17]

It is an example of balanced force.

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A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

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3 years ago

What does it mean for something to have charge?

Lana71 [14]

The property of matter that is responsible for electrical phenomena, existing in a positive or negative form.

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2 years ago

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A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

$\lambda = \frac{c}{f}$

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

I hope it helps you!

7 0

3 years ago