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2 years ago

Which of Newton's laws accounts for the following statement? "A force cannot act alone." first law second law third law

Physics

2 answers:

enot [183]2 years ago

6 0

Newton's laws A force cannot act alone is the THIRD LAW!

ki77a [65]2 years ago

4 0

Answer:

"A force cannot act alone" is a statement that belongs to Newton's Third Law, because the forces act in pairs (action and reaction).

Explanation:

Newton's First Law, also called the Law of inertia, indicates that "The whole body perseveres in its state of rest or uniform rectilinear motion, unless the sea is forced to change its state by forces printed on it." This means that for a body to leave its state of rest or the motion, it is necessary for a force to act on it.

Newton's Second Law, also called the Fundamental Law of Dynamics, indicates that: "The change of motion is directly proportional to the printed driving force and occurs along a straight line along which that force is printed." This means that the acceleration of a moving object depends on the amount of force that is applied at a given time, in order to modify its trajectory or speed.

Newton's Third Law is also known as the Principle of action and reaction. This law indicates that “Every action has an equal reaction but in the opposite direction” This means that when a body exerts a force (action) on another, it responds with a force of equal magnitude although opposite direction (reaction). Then this law explains the forces that always occur in the form of peers: an action and a reaction.

Finally <u><em>"A force cannot act alone" is a statement that belongs to Newton's Third Law, because the forces act in pairs (action and reaction).</em></u>

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A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im

geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

Em₀ = Em_f

½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

v_f ² = 20² + 2 9.8 (10 -15)

v_f = √302

v_f = 17.4 m / s

8 0

2 years ago

A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei

QveST [7]

gravitational potential energy is given by formula

$U = mgh$

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

$\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}$

$\frac{U_2}{U_1} =\frac{m_2}{m_1}$

given that

$m_2 = 4 m_1$

now we have

$\frac{U_2}{U_1} = 4$

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2 years ago

Which of the choices describes an action–reaction force pair for a space station containing astronauts in orbit about the earth?

11111nata11111 [884]

Answer:

(D) The weight of the space station and the gravitational force of the space station on the earth.

Explanation:

In both A and B , both the forces act in the same direction ( downwards ) , so they can not be action- reaction force .

In the option C , weight of a astronaut can only be reaction force of gravitational force exerted on the earth by astronaut. Both astronaut and the earth pull each other with equal and opposite force. So option D is correct.

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2 years ago

An object weighs 200 newtons at a distance of 100 kilometers above the center of a small uniform planet. how much will the objec

disa [49]

Since the law of gravitation is an inverse square law if you quadruple the radius the f will drop by a factor of 16 SO the object would weigh 200/16 = 12.5N

In other words, as the distance, or radius, quadruples the weight becomes 1/16 of the original weight. Just plug in 4 for r and when you square it you get 16. The numerator is 1 so that is how the weight becomes 1/16.

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2 years ago

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pantera1 [17]

The motion of the buoy consists of two independent motions on the horizontal and vertical axis.

On the horizontal axis, the motion of the buoy is a uniform motion with constant speed $v=50 m/s$ . On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ . The vertical position of the buoy at time t is given by
$y(t)=h- \frac{1}{2}gt^2$
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m$

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1 year ago