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noname [10]
2 years ago
5

What is the term for the principal that an object in motion remains in motion until it is met by an outside force?

Physics
1 answer:
Papessa [141]2 years ago
7 0

The term for the principal that an object in motion remains in motion until it is met by an outside force is called inertia

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force .This principal is named as inertia .

Every matter have inertia , a property to remain in that very position until unless it is been forced on them to change their position .

Inertia is a property of matter that causes it to resist changes in velocity (speed and/or direction). According to Newton's first law of motion, an object with a given velocity maintains that velocity unless acted on by an external force. Inertia is the property of matter that makes this law hold true.

To learn more about inertia here

brainly.com/question/1358512?referrer=searchResults

#SPJ4

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The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass
Triss [41]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • spring constant of the spring attached to the input piston, k=1600\ N.m^{-1}
  • mass subjected to the output plunger, m=40\ kg

<u>Now, the force due to the mass:</u>

F=m.g

F=40\times 9.8

F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

8 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

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An object moving at 38m/s takes 4s to come to a stop. What is the object’s acceleration?
Rudik [331]
- 9.5 m/s^2

use the SUVAT method 

0 = 38 + 4a
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A girl runs and cover a distance of 200m in 20 secends.The speed of the girl is ................ m/s.
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Speed = distance/time

Speed = (200m)/(20s)

Speed = 10 m/s

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