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kari74 [83]
3 years ago
6

Why the dust particles undergoes small random movements? Plz help!!?

Physics
1 answer:
Rashid [163]3 years ago
8 0
Brownian motion<span> or pedesis is the </span>random motion<span> of particles suspended in a fluid </span>
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A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before
Airida [17]

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so  W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J


5 0
4 years ago
Josh is designing a lab to separate a mixture of sulfur powder and iron filings. He looks at all the ways to separate a mixture
Stolb23 [73]

Answer:

D) magnetic attraction

Explanation:

Mixture consists of sulfur powder and iron filings . The former is non magnetic and the later is magnetic material . The best way to separate them is to use a magnet . This method is called magnetic separation method . A magnet will attract the iron filling but the sulfur will remain un-attracted . Thus, the two will get separated from each other.

6 0
3 years ago
Read 2 more answers
A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the
LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

  • 1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

  • f = Focal length of the ornament
  • u = image distance
  • v = object distance.

make u the subject of the equation

  • u = fv/(f+v)................ Equation 2

From the question,

Given:

  • f = 8/2 = 4 cm
  • v = 12 cm

Substitute these values into equation 2

  • u = (12×4)/(12+4)
  • u = 48/16
  • u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

  • M = u/v.................. Equation 3

Where

  • M = magnification of the ornament.

Substitute these values above into equation 3

  • M = 3/12
  • M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0
3 years ago
For each 1 degree Celsius that temperature incileases, the speed of<br> sound ___ by 0.6 m/s.
Levart [38]

Answer:

A: For each 1 degree Celsius that temperature decreases, the speed of sound decreases by 0.6 m/s. So sound travels through dry, -20 °C air at a speed of 319 m/s

Explanation:

6 0
3 years ago
A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl
Triss [41]

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

8 0
3 years ago
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