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anygoal [31]
2 years ago
10

What is the weight of an object(on earth) that has a mass of 50Kg (g earth=10N/Kg)

Physics
1 answer:
kumpel [21]2 years ago
4 0
W=mg
Therefore W is 50by 10 thus 500N
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A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho
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Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

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Where do tadpoles live? a. on land c. both on land and water b. in water d. none of the above
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Tadpoles live in water
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Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat
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Answer:

475

Explanation:

Cori does not exert any more force than 475 J, so 475 is the answer.

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2 years ago
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Why is Pluto no longer a planet?
Step2247 [10]
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If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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