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guapka [62]
3 years ago
10

A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is

done?
Physics
1 answer:
Tpy6a [65]3 years ago
5 0

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

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How do leptons differ from hadrons?
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The distance between two corresponding parts of a wave is its
oksano4ka [1.4K]

Answer:

wavelenght

Explanation:

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Read 2 more answers
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
4 years ago
For a 50 kV anode voltage, what is the maximum photon energy of the x-ray radiation?
V125BC [204]

Answer:

The energy of photon, E=8\times 10^{-15}\ J

Explanation:

It is given that,

Voltage of anode, V=50\ kV=50\times 10^3\ V=5\times 10^4\ V

We need to find the maximum energy of the photon of the x- ray radiation. The energy required to raise an electron through one volt is called electron volt.

E=eV

e is charge of electron

E=1.6\times 10^{-19}\times 5\times 10^4

E=8\times 10^{-15}\ J

So, the maximum energy of the x- ray radiation is 8\times 10^{-15}\ J. Hence, this is the required solution.

6 0
3 years ago
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