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makkiz [27]
3 years ago
14

An object was dropped from the plane with 1000 km above the ground with a mass of 300 kg. Find the acceleration of the object ea

ch second until it reaches the ground.
Physics
1 answer:
Oxana [17]3 years ago
8 0

Answer:

9.8m/s^2

Explanation:

The acceleration due to gravity on earth is 9.8m/s^2 regardless of the mass of the object. This means, in the absence of air resistance, both a 3000 kg object and a 300 kg will accelerate at the same rate of 9.8m/s^2 towards  earth.

The reason the acceleration due to gravity is the same for all masses is that by grace of nature, the inertial mass m of the object from F= ma  equals the gravitational mass m from F =G \dfrac{mM}{R^2}, resulting

ma=G\dfrac{mM}{R^2}

\boxed{a=G\dfrac{M}{R^2}}

which is the acceleration that only depends on the mass of earth and its radius, and not on the mass of the object.

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Once the virus gets into the host's body, it enters the nearest cell.Once it gets to the nucleus it turns that cell into a virus-production cell and once the immune system detects those viruses it takes action.
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Which instrument would make rice vibrate easier, a tuba or a flute? Explain why. Hint: think about the difference between high a
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I assume the higher notes would make the rice vibrate more easily, so a flute.

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3 years ago
Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?
lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

  • F₁ = 225.4 N
  • F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

            ∑ F = 0

            ∑ τ = 0

           

Where F are the forces and τ the torques.

The torque  is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

           F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

         F₂ 2 - W₁ 1 - W₂ 1.5 = 0\frac{W_1  \ 1 + W_2 \ 1.5}{2}

Let's calculate F₂

         F₂ = \frac{W_1 \ 1 + W_2 \ 1.5 }{2}  

         F₂ = (m g + M g 1.5)/ 2

         F₂ = \frac{(12 + 68 \ 1.5 ) \  9.8}{2}  

         F₂ = 558.6 N

We substitute in the translational equilibrium equation.

         F₁ = W₁ + W₂ - F₂

         F₁ = (m + M) g - F₂

         F₁ = (12 +68) 9.8 - 558.6

         F₁ = 225.4 N

In conclusion using the equilibrium conditions we can find the forces of the balance are:

  • F₁ = 225.4 N
  • F2 = 558.6 N

Learn more here:  brainly.com/question/12830892

5 0
2 years ago
D<br> Question 14<br> 2 pts<br> The universe could be considered an isolated system because
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Answer:

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Explanation:

5 0
3 years ago
How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent
Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0
3 years ago
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