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4vir4ik [10]
2 years ago
7

What is point c called?

Physics
2 answers:
FromTheMoon [43]2 years ago
5 0
Then every line segment has one and only one mid-point.

(っ^▿^)
zmey [24]2 years ago
3 0
Point C is always line segment
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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
2 years ago
Is frequent anger a problem for you? If so, what strategies could you try to<br> overcome it?
erastova [34]

Answer:

I tend to slap thicc trees a#ss

Explanation:

3 0
3 years ago
A car moved 120km to the north. what is its displacement?
inn [45]
120Km. They are the same
5 0
3 years ago
A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi
oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775° 

 

<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>

F net = 49N 

<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>

<span>Work = 980 J </span>

4 0
3 years ago
Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.
sp2606 [1]

Answer:

6J

Explanation:

Given parameters:

Mass of fish  = 1kg

Velocity  = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

       K.E  = \frac{1}{2}  m v²  

Now, insert the parameters and solve;

  K.E  =  \frac{1}{2}  x 1 x 12  = 6J

The change in kinetic energy is 6J

4 0
2 years ago
Read 2 more answers
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