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lesya [120]
3 years ago
10

70 kg to mg i need to show the work of how i did it

Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

70000000 mg

Explanation:

1 gram (kg) is equal to 1000000 milligrams (mg).

1 kg = 106 mg = 1000000 m

7 kg × 1000000

= 7000000 mg

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A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal
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Answer:

Given that,  

Speed = v  

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We have to ascertain the precise speed  

Utilizing equation of speed  

Where, v = velocity/speed

r = radius  

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The precise speed is characterized by the speed of turn.  

The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.  

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3 years ago
A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150
vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

  • Mass of the box ( m ) = 37.5 kg.
  • Displacement made by box ( x ) = 4.05 m.
  • Horizontal force ( F ) = 150 N.
  • The co-efficient of friction between box and floor ( μ ) = 0.3
  • Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0
3 years ago
Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
Anna11 [10]

Kepler’s third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496 × 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/( 86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The closest answer is 1.99 × 10^30

(it may vary a little with rounding – the difference is less than 1%)

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