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zubka84 [21]
3 years ago
5

What is the acceleration of an object with mass of 42.6 kg when an unbalanced force of 112 N is applied to it

Physics
2 answers:
a_sh-v [17]3 years ago
8 0

Answer:

2.63m/s^2

Explanation:

To solve this problem we use Newton's second law, which establishes a relationship between mass, acceleration and force applied to an object.

F=ma

Where F  is the force, m is the mass and a is the acceleration. Because they ask us for acceleration, we clear for  a:

a=\frac{F}{m}

And since we have the following information:

m=42.6kg and F=112N

the acceleration will be:

a=\frac{112N}{42.6kg}

a=2.629m/s^2≈ 2.63m/s^2

The answer is that the acceleration of the object is 2.63m/s^2

lord [1]3 years ago
7 0
The answer is 2.63m/s^2! You use the formula F=ma, 112 = 42.6(a), a= 2.63m/s^2.
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A star produces 2x10^26 watts. how much energy does it lose every minutes
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Answer:

Energy loss per minute will be 120\times 10^{26}j

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We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

We multiply with 60 we have to calculate energy loss per minute

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The amount of work done against friction to slide a box in a straight line across a uniform, horizontal floor depends most on th
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An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?
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We know, weight = mass * gravity 
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r²   [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)²   [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
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In short, Your Answer would be 4 Newtons

Hope this helps!
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iragen [17]

The Moment of Inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

I = I_{D} - I_{H} (1)

Where:

  • I_{D} - Moment of inertia of the Disk.
  • I_{H} - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right) (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole (m):

\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}

m = \frac{1}{2}\cdot M

And the resulting equation is:

I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)

I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}

I = \frac{15}{32}\cdot M\cdot R^{2}

The moment of inertia of the Disc is represented by I = \frac{15}{32}\cdot M\cdot R^{2}. (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

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