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sergiy2304 [10]
1 year ago
14

Question 9 of 10

Physics
1 answer:
romanna [79]1 year ago
5 0

A atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

<h3>What are nuclear forces?</h3>

The expression 'nuclear forces' makes reference to the attraction between different subatomic particles in a given nucleus of a particular atom, which represents the strongest force in the universe.

Subatomic particles include positively charged particles (i.e., protons), neutrally charged particles called neutrons, and electrically negative charged particles called electrons (e-).

In conclusion, a atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

Learn more about nuclear forces here:

brainly.com/question/16959709

#SPJ1

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the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

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3 years ago
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MakcuM [25]

Answer:C

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3 years ago
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the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass
DiKsa [7]
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3 years ago
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A Person whose weight is 5.20 x 10^2 N is being pulledup
Gnoma [55]

Answer:

Explanation:

Given

Weight of Person W=5.20\times 10^{2} N

Cave is h=35.1 m deep

Breaking stress T=569 N

Net Force on Person

F_{net}=569-520=49 N

a_{net}=\frac{F_{net}}{\frac{W}{g}}

a_{net}=\frac{49}{\frac{520}{9.8}}

a_{net}=0.923 m/s^2

The shortest time such that the person can be taken out of cave

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

where

h=distance moved

t=time

a=acceleration

35.1=0+\frac{1}{2}(0.923)(t)^2

t^2=76.05

t=\sqrt{76.05}

t=8.72\ s                    

6 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
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