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Alexeev081 [22]
3 years ago
11

What is the SI unit of force?​

Physics
1 answer:
seraphim [82]3 years ago
5 0

The SI unit of force is the<em> Newton</em> .

1 Newton is the force that accelerates 1 kilogram of mass at the rate of 1 meter per second-squared.

1 pound of force is roughly 4.45 Newtons of force.

So 1 Newton of force is roughly 3.6 ounces of force.

1 kilogram of mass weighs about 9.81 Newtons on Earth.  That's the same as 2.205 pounds.

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A bus driver heads south with a steady speed of
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ANS

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What's the formula to calculate the radius of pendulum bob ​
Svetradugi [14.3K]

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7 0
3 years ago
A 2.964 kilogram truck strikes a fence at 7.00 meters/second and comes to
Makovka662 [10]

10.92N

Explanation:

Given parameters:

Mass of truck = 2.964kg

Velocity of truck = 7m/s

Time taken = 1.9s

Unknown:

Average force on the car = ?

Solution:

According to newton's third law of motion "action and reaction are equal and opposite".

The force with which the truck struck the fence is the same as the force the fence acted on the truck with but in another direction.

 From newton's second law:

      Force  = mass x acceleration

      We know that acceleration is the change in velocity with time;

   acceleration = \frac{change in velocity }{time }

   Force =  mass x  \frac{change in velocity }{time }

  Force = \frac{2.964 x 7}{1.9} = 10.92N

learn more:

Newton's laws brainly.com/question/11411375

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3 0
3 years ago
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
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