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Harrizon [31]
3 years ago
5

An alpha particle (which has two protons) is sent directly toward a target nucleus with 90 protons. the alpha particle has a kin

etic energy of 0.45 pj when far away. what is the least center-to-center distance that the alpha particle will be from the target nucleus, assuming that the nucleus does not move?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0
It does not move because of the sun the sun has no energy.
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The fulcrum is between the effort and the load
Inessa05 [86]

Class 1 lever

Explanation:

In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.

  • The effort is the force input and the load is the force output
  • The fulcrum is a hinge between the load and effort.
  • Movement of the effort and load are in opposite directions.
  • There are other classes of lever like the class 2 and 3.
  • They all have different load, fulcrum and effort configurations

learn more:

Load related problems brainly.com/question/9202964

Torque brainly.com/question/5352966

#learnwithBrainly

3 0
3 years ago
Can someone please help
Maru [420]

Answer: D

Explanation: because doing a yoga desk program is physical activity, 10k steps is pysical activity, riding a bike or walking/running is also physical activity. so it should be D, all of the above.

5 0
3 years ago
Hope you all doing okay
Vsevolod [243]

Answer:Thank you

Explanation:

8 0
1 year ago
Plz help me with this activity :(
Marrrta [24]
As mass increases, the potential energy also increases, I hope that helped :)

8 0
3 years ago
A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a
frosja888 [35]

Answer:

a) Yes

b) 7 rad/s

c) 0.01034 J

Explanation:

a)

Yes the angular momentum of the block is conserved since the net torque on the block is zero.

b)

m = mass of the block = 0.0250 kg

w₀ = initial angular speed before puling the cord = 1.75 rad/s

r₀ = initial radius before puling the cord = 0.3 m

w = final angular speed after puling the cord = ?

r = final  radius after puling the cord = 0.15 m

Using conservation of angular momentum

m r₀² w₀ = m r² w

r₀² w₀ = r² w

(0.3)² (1.75) = (0.15)² w

w = 7 rad/s

c)

Change in kinetic energy is given as

ΔKE = (0.5) (m r² w² - m r₀² w₀²)

ΔKE = (0.5) ((0.025) (0.15)² (7)² - (0.025) (0.3)² (1.75)²)

ΔKE = 0.01034 J

6 0
3 years ago
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