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daser333 [38]
3 years ago
10

Newton's third law is applicable only to objects at rest. True False

Physics
2 answers:
Wittaler [7]3 years ago
8 0
Newton's third law applies even when objects do not move. Here a gymnast pushes downward on the bars. The bars push back on the gymnast with an equal force. an action-reaction pair is because one of the objects is often much more massive and appears to remain motionless when a force acts on it.
DiKsa [7]3 years ago
5 0

Answer:

False

Explanation:

Newton's third law - For every action there's an equal and opposite reaction, action and reaction act on different objects

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Describe three physical changes that occur in nature?
Alex Ar [27]

Answer: Physical changes in nature could then be erosion in a mountain, the melting of snow, and a river freezing over from the cold. Since none of these changes affect the chemical composition of the mountain, the snow, or the river, they are physical changes.

Explanation:

8 0
3 years ago
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Which statement can be supported by using a position-time grap
Phoenix [80]

Answer:

A negative slope results when an individual is moving away

Explanation:

6 0
3 years ago
when a stationary rugby ball is kicked, it is contact with a player's about for 0.05 s. during this short time, the ball acceler
Ad libitum [116K]

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

3 0
3 years ago
Starting from rest, an intern pushes a 42-kggurney 12 m down the hall with a constant force of 80 N directed downward at an angl
DerKrebs [107]

Answer:

A. The work done by the intern is 792 J.

B. The velocity of the gurney when it has moved 12 m is 6.1 m/s.

C. The 12-m journey takes 3.8 s.

Explanation:

Hi there!

Please see the attached figure for a description of the situation.

<u>Part A: </u>

Work is done by a force when it is applied in the direction of the displacement or against it. In this case, the only force applied in the direction of displacement is the horizontal component of the force applied by the intern.

By trigonometry, the horizontal component of the force is calculated as follows:

cos θ = adjacent/hypotenuse

Looking at the figure, you can notice that the applied force, F, is the hypotenuse of a right triangle and the horizontal component, Fx, is the adjacent side:

cos θ = Fx / F  

Fx = F · cos θ

Fx = 80 N · cos 35°

Fx = 66 N

Now we can calculate the work (W) done by this force:

W = Fx · x

Where x is the displacement:

W = 66 N · 12 m = 792 J

The work done by the intern is 792 J.

<u>Part B:</u>

Applying the work-energy theorem, the work done is equal to the change in kinetic energy:

W = final kinetic energy - initial kinetic energy

Since the gurney starts from rest, the initial kinetic energy is zero. Then:

W = final kinetic energy

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the gurney.

v = velocity.

KE = 792 J

792 J = 1/2 · 42 kg · v²

v²= 2· 792 J / 42 kg

v = 6.1 m/s

The velocity of the gurney when it has moved 12 m is 6.1 m/s.

<u>Part C:</u>

First, let´s find the acceleration of the gurney:

Fx = m · a

Fx/m = a

66N / 42 kg = a

a = 1.6 m/s²

Now using the equation of velocity, let´s find the time at which the gurney has a velocity of 6.1 m/s:

v = v0 + a · t

Where:

v = velocity at time t.

v0 = initial velocity.

a = accleration.

t = time.

v = v0 + a · t

6.1 m/s = 0 + 1.6 m/s² · t

t = 6.1 m/s / 1.6 m/s²

t = 3.8 s

The 12-m journey takes 3.8 s.

3 0
3 years ago
The new horizons spacecraft, launched in 2006, spent 9.5 years on its journey to pluto. it derives its electric power from the h
natta225 [31]

The total thermal power generated by this plutonium source is determined as 5.65 J/s.

<h3>Total thermal power generated</h3>

The total thermal power generated by this plutonium source is determined as follows;

ΔP = Eλ

where;

  • E is the energy of the particles = 5.6 mev
  • λ is the activity of the plutonium, which should be = 6.3 x 10¹⁵ s⁻¹

ΔP = (5.6 mev) x (6.3 x 10¹⁵ s⁻¹)

ΔP = (5.6 x 10³ x 1.6 x 10⁻¹⁹ J) x (6.3 x 10¹⁵ s⁻¹)

ΔP = 5.65 J/s

Learn more about thermal power here: brainly.com/question/7541718

#SPJ1

7 0
3 years ago
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