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labwork [276]
3 years ago
11

I am the particle that is so small that scientists treat me as though I have no mass. Scientists also figure that I probably spe

nd most of my time in the cloud outside the nucleus. Give my name? can yall plz help me I'm in 8th grade and I need to know what this is because this is due tomorrow ​
Physics
1 answer:
tatyana61 [14]3 years ago
8 0

Answer:

its electrons

Explanation:

only electrons stays outside the nucleus unlike protons and neutrons and it has little to no mass

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ball is dropped from rest and hits the ground 1.1 seconds later. What height was the ball released from? ​
Aleonysh [2.5K]
Height = 1/2 * 9.8 * (1.1^2)
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3 years ago
Which of the following is a category of mechanical wave?
givi [52]

Answer:

a

because the mechanical wave is when it goes over and over again

8 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
2 years ago
The distance between two planets is 1600 km. How much time would the light
Snowcat [4.5K]

Answer:

5.33*10^-3 seconds

Explanation:

c = d/t

c = speed of light constant (3.0*10^5 km/s)

d = distance (1600 km)

t = ?

3.0*10^5 = 1600/t

t = 1600/3.0*10^5

t = 5.33*10^-3 seconds

I hope this helped! :)

6 0
3 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
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