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uranmaximum [27]
1 year ago
6

what is the magnitude of a force?(1 point) responses the motion of the force the motion of the force the direction of the force

the direction of the force the type of force the type of force the amount of force the amount of force
Physics
1 answer:
mel-nik [20]1 year ago
6 0

The term "unbalanced force" in Newton's first law refers to a force that is not entirely counterbalanced (or cancelled) by the other independent forces. Basically, magnitude can be thought of as simply the "value" or "amount" of any physical quantity. It is a scalar quantity at all times.

Force has both a magnitude and a direction because it is a vector quantity. The amount that encapsulates the force's strength is known as its magnitude.

Consider the following scenario: the force is 10 N in the east. The direction is indicated by "towards east," while the force is indicated by "10."

Unless acted upon by an unbalanced force, an object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and direction.

The term "unbalanced force" in Newton's first law refers to a force that is not entirely counterbalanced (or cancelled) by the other independent forces. An uneven force exists if either all the vertical forces (up and down) or all the horizontal forces do not cancel each other. Looking at the free-body diagram for a particular situation makes it easy to recognize whether an unbalanced force is present.

To know more about unbalanced force click on the link:

brainly.com/question/19054208

#SPJ4

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A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude
irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b.\mid A\mid=46 m

\theta=20^{\circ} below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=x=360-20=340^{\circ}

x-component of vector A=A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24

y-Component of vector A=A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis=x'=42^{\circ}

x-component of vector B=B_x=86cos42=63.64

y-Component of vector B=B_y=86sin42=57.62

Vector A=A_xi+A_yj=43.24i-15.64j

Vector B=B_xi+B_yj=63.64i+57.62j

Vector C=A+B

Substitute the values

C=43.24i-15.64j+63.64i+57.62j

C=106.88i+41.98j

c.Direction=\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}

The direction of the vector C=21.5 degree

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Answer:

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Explanation:

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