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IRINA_888 [86]
3 years ago
6

Newton’s first law relates motion to balanced and unbalanced forces.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
8 0
True, an object at rest stays and rest and an object in motion stays in motion
gulaghasi [49]3 years ago
8 0

YES !  

True and certainly.  

Uh huh, uh huh.

That's a fact.  

More intense veracity is seldom encountered.

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
the value of acceleration due to gravity option A is same on equator and poles option B is list on poles option C is list on equ
stiks02 [169]

Answer:

D. Increases from pole to equator

Explanation:

I majored in Science

8 0
3 years ago
Describe the motion of an unanchored rowboat when a water wave passes
USPshnik [31]
It mimics the movement of the waves 
6 0
3 years ago
Read 2 more answers
what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?
nignag [31]

Using Newton's second law of motion:

F=ma ;  [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)


Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

8 0
3 years ago
A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as
malfutka [58]

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2\pi \sqrt{0.5*9.8*0.3}

T=1.2042s

The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

#SPJ1

6 0
1 year ago
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