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3 years ago

Action and reaction forces will cancel because they are pushing on the same object toward each other. True or False

Physics

2 answers:

sweet-ann [11.9K]3 years ago

5 0

Answer:

False

Explanation:

Newton's third law states that for every action there is an equal and opposite reaction. The action-reaction pair act on different bodies. Thus, they do not cancel each other.

For example, in order to walk, we push the floor backwards. This is action force which the foot applies to the floor. The floor pushes the foot forward. This is the reaction force. The amount of force is equal but the direction is opposite. Also, the bodies on which action force and reaction force are acting are different.

Hence, the given statement is false.

vampirchik [111]3 years ago

4 0

Sometimes. If the forces are equal, they will cancel out and the net force will be zero. But if the forces are not equal, the object will move away from the stronger force because there is still a net force.

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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

4 0

4 years ago

1.5 x 10^3 standard notation

poizon [28]

Answer:

1500

Explanation:

3 0

3 years ago

Choose the correct statement regarding projectile motion in the absence of air resistance. Assume the object is at sea level mov

pishuonlain [190]

Answer:

Explanation:

The magnitude of the acceleration in the x direction is always zero: TRUE.

At the apex, in the y direction the velocity is zero and the accelration is positive. TRUE.

At the apex, in the x direction, the velocity is zero and the acceleration is zero. FALSE. The accelration is zero, but the velocity is the same it had when it was shot.

The magnitude of the velocity in the y direction is always constant. FALSE, it's subject to gravity and it's velocity varies as $v=v_0 - 9.81ms^{-2} t$

At the apex, in the Y direction, the velocity is negative and the acceleration is zero. FALSE. Velocity is zero, Acceleration is $9.81 ms^{-2}$ , towards the negative y axis