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Helga [31]
3 years ago
9

A student is helping her teacher move a 9.5 kg box of books. what net sideways force must she exert on the box to slide it acros

s the floor so that it accelerates at 1.0 m/^2?
Physics
1 answer:
Tanzania [10]3 years ago
7 0

Mass of the box which contains books (m) = 9.5 kg

Acceleration required (a) = 1 m/s^2

Let the force applied by the teacher and student be F.

The only force acting in the horizontal direction is F and it will produce an acceleration of 1 m/s^2

According to Newton's Seconds law:

F(net) = ma

F = ma

F = 9.5 × 1

F = 9.5 Newtons

Hence, the sideways force applied by teacher and the students is equal to 9.5 Newtons

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Answer:

battery

Explanation:

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A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a
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For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

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3 years ago
What is the average speed for the entire graph?
Maru [420]

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7 0
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You might say that this experiment was an attempt to build a scale, and then calibrate it against a scale that we trust (the ele
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No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

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Random errors can be minimized by taking multiple readings and averaging the results.

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3 0
3 years ago
Read 2 more answers
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and
Alexxandr [17]

Answer:

a

  F  =326.7 \ N

b

  M  = 6

Explanation:

From the question we are told that

          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

6 0
3 years ago
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