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Salsk061 [2.6K]
3 years ago
14

A rock has a mass of 3.1 kg. What is its weight on earth

Physics
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

W = 30.38 N

Explanation:

Given that,

Mass of a rock, m = 3.1 kg

We need to find the weight of the rock on the surface of Earth. Weight of an object is given by :

W = mg

g is the acceleration due to gravity, g = 9.8 m/s²

W = 3.1 kg × 9.8 m/s²

= 30.38 N

So, the weight of the rock on the Earth is 30.38 N.

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The answer is B. Scientific Law

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If i click on a mouse does that show energy transfer?
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To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

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    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

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       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

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3 years ago
How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
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The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
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