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wlad13 [49]
3 years ago
14

How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at a constant velocity 0.3 meter from a ta

ble?
Physics
1 answer:
pychu [463]3 years ago
5 0

Answer:

0.2943 Nm

Explanation:

Work done is given a the product of force and diatance moved and expressed by the formula

W=Fd

Here W represent work, F is applied force and d is perpendicular distance

Also, we know that F=mg where m is the mass of an object and g is acceleration due to gravity. Substituting this back into the initial equation then

W=mgd

Taking acceleration due to gravity as 9.81 m/s2 and substituting mass with 0.1 kg and distance with 0.3 m then

W=0.1*9.81*0.3=0.2943 Nm

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The  kinetic energy is KE  =  7.59  *10^{10} \  J

Explanation:

From the question we are told that

       The  radius of the orbit is  r =  2.3 *10^{4} \ km  = 2.3  *10^{7} \ m

       The gravitational force is  F_g  = 6600 \ N

The kinetic energy of the satellite is mathematically represented as

       KE  =  \frac{1}{2} * mv^2

where v is the speed of the satellite which is mathematically represented as

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=>  v^2  =  \frac{GM }{r}

substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

         KE  =  7.59  *10^{10} \  J

 

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