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Rudiy27
2 years ago
10

A 200-kg object and a 500-kg object are separated by 4.00 m. (a) find the net gravitational force exerted by these objects on a

50.0-kg object placed midway between them. (b) at what position (other than an infinitely remote one) can the 50.0-kg object be placed so as to experience a net force of zero from the other two objects?

Physics
1 answer:
pantera1 [17]2 years ago
5 0
Check the attached file for the solution for this problem.

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Explain how energy is conserved when nuclear fission or fusion occurs
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The mass lost in the nuclear reaction is all converted to energy.
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If the length of the ramp is 2m and the height of the ramp is 1m , what is the mechanical advantage?
slamgirl [31]

Answer:

Mechanical Advantage is 2

Explanation:

M.A = length of ramp / height of ramp

= 2/1

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5 0
3 years ago
How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number mℓ = −1?
LekaFEV [45]

Answer:

6 electrons

Explanation:

To solve this question lets determine the possible quantum numbers for the principal quantum number n = 4.

For the quantum number l which describes the shape of the orbital, we have the possible values : 0 to n-1.

Thus, for n = 4 the l can assume the values 0, 1, 2, 3 ( 4 possible shapes )

For the angular quantum number, ml, which tell us the orientation in space,we have the values - l to + l.

So lets determine the number of orbials which can have the values -l for n=4

l = 0   ml = 0

l=  1    ml = -1,0,1

l= 2    ml = -2, -1, 0, 1, 2

l= 3    ml = -3,-2,-1,0,1,2,3

So we have three orbital with ml = - 1 and from Pauli´s exclusion principle we can have up two electrons in each orbital. Thus for n= 4 we can have up to 6 electrons with ml = -1

3 0
3 years ago
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

6 0
3 years ago
Huryyy
zmey [24]
The transfer is perpendicular

Surprised
 I knew this<span />
3 0
3 years ago
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