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noname [10]
2 years ago
14

A rightward force of 4.0 N is exerted upon an object for a distance of 3.0 meters.

Physics
1 answer:
julia-pushkina [17]2 years ago
8 0

Answer:

W = 12 J

Explanation:

Given that,

Force, F = 4 N

The object moves in rightward direction for a distance of 3 m.

Work done on the object is given by :

W=F\times d\\\\=4\ N\times 3\ m\\\\=12\ J

So, the work done on the object is 12 J.

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Wolfgang pauli hypothesized an exclusion principle. This principle says two electrons in an atom cannot have the same what?.
tatiyna

No two electrons in an atom or molecule may have the same four electronic quantum numbers, according to the Pauli Exclusion Principle. Only two electrons can fit into an orbital at a time, hence they must have opposing spins.

<h3>What is Pauli's exclusion principle ?</h3>

According to Pauli's Exclusion Principle, no two electrons in the same atom can have values for all four of their quantum numbers that are exactly the same. In other words, two electrons in the same orbital must have opposing spins and no more than two electrons can occupy the same orbital.

  • The reason it is known as the exclusion principle is because it states that all other electrons in an atom are excluded if one electron in the atom has the same specific values for all four quantum numbers.

Learn more about Pauli's exclusion principle here:

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3 0
1 year ago
Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by ma
Tamiku [17]

Answer:

Explanation:

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8 0
3 years ago
How can I solve this?
inysia [295]
There is no equation here
3 0
3 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

   P₁ = m v

   P₁ = 0.0575 x 26.7

   P₁ = 1.535 kg.m/s

b) momentum of ball after collision

   P₂ = m v'

   P₂ = 0.0575 x (-29.5)

   P₂ = -1.696 kg.m/s

c) change in momentum

    Δ P = P₂ - P₁

    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

  initial speed of racket = 0 m/s

  M u + m v = Mu' + m v

  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
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