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iogann1982 [59]
4 years ago
14

Which event is an example of condensation?

Physics
1 answer:
Sonbull [250]4 years ago
5 0

Answer:

Fog forms in a valley

Explanation: that’s the only reasonable response

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What is a
lawyer [7]

Answer:

A. They are a solid.

Explanation:

Hope this helps. . .

uwu

3 0
3 years ago
Read 2 more answers
I need help with one through six please
dybincka [34]

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

 V = \frac{50}{10}

 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

 A = \frac{30-15}{3}

 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

7 0
2 years ago
Jennifer's cat is stuck in a
Lelu [443]

The speed with which Jennifer climb up the tree is 0.45 m/s

The distance is 10m

Time is 22 seconds

Required speed is 10/22=0.45 m/s

  • Distance is the total length travelled by an object.
  • Speed is distance upon time.
  • It is a scalar quantity and doesn't has magnitude and direction.
  • Distance traveled is the total length of the path traveled between two positions
  • Distance is the total movement of an object without any regard to direction.
  • Distance is the total movement of an object without any regard to direction.

To learn more about distance visit at :

brainly.com/question/15256256

#SPJ9

5 0
2 years ago
In the Bohr’s model of the hydrogen atom, the electron moves in a circular orbit of radius 5.041 × 10−11 m around the proton. As
Ilia_Sergeevich [38]

Answer:

The orbital speed of the electron is 2.296 x 10⁶ m/s

Explanation:

Given;

radius of the circular orbit, r = 5.041 × 10⁻¹¹ m

In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;

V = \frac{nh}{2\pi mr}

where;

h is Planck's constant

m is mass of electron

r is the radius of the circular orbit

n is the energy level of hydrogen in ground state

Substitute in these values and solve for V

V = \frac{nh}{2\pi mr} = \frac{1*6.626*10^{-34}}{2\pi *9.11 *10^{-31}*5.041*10^{-11}}\\\\V = 2.296*10^6  \ m/s

Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s

4 0
4 years ago
A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m lon
ch4aika [34]

To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is

KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f

KE_i = 0, since initially cylinder was at rest

PE_f = 0 since at the ground potential energy is zero

The mathematical values are,

mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2

Here,

m = mass

g= Gravity

h = Height

V = Velocity

I = \frac{mr^2}{2} moment of Inertia in terms of its mass and radius

\omega = \frac{V}{r} Angular velocity in terms of tangential velocity and its radius

Replacing the values we have that

mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2

gh = \frac{v^2}{2}+\frac{v^2}{4}

v = \sqrt{\frac{4gh}{3}}

From trigonometry the vertical height of inclined plane is the length of this plane for sin\theta, then

h = 2.00*sin 25

h = 0.845 m

Replacing,

v = \sqrt{\frac{4(9.8)(0.845)}{3}}

V = 3.32 m/s

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s

6 0
3 years ago
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