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stiv31 [10]
3 years ago
5

John McRighteous has a mass of 90kg. Calculate johns weight.

Physics
2 answers:
Natalija [7]3 years ago
3 0

Answer: W=M*gravity. 90 * 9.8=882N. That is the weight on Earth.

Explanation:

Calculating the weight is W=M*gravity. 90 * 9.8=882N. That is the weight on Earth.

forsale [732]3 years ago
3 0

Answer:

198.4 lbs

Explanation:

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A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.
rodikova [14]

Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²

Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

3 0
3 years ago
Increasing the number of loops in a electromagnet or solenoid will cause it to be stronger
Zielflug [23.3K]
True

The electromagnet will become stronger if we add more coils because there are more field lines in a loop then there is in a straight piece of wire. In a solenoid there are a lot of loops and they are concentrated in the middle, as more loops are added the field lines get larger, therefore making the electromagnet stronger.
5 0
3 years ago
Metamorphic rocks may generally be distinguished from igneous rocks because
nlexa [21]
The grains in igneous rocks are interlocking but the Metamorphic rocks are not.
6 0
3 years ago
An object with a mass of 1.5 kg accelerates 10.0 m/s2? when an unknown force is applied to it. What is the amount of the force?
Svetlanka [38]

Answer:

15N

Explanation:

f=ma

1.5kg x 10m/so

f= 15N

7 0
3 years ago
A +5.00 pC charge is located on a sheet of paper.
emmainna [20.7K]

Answer:

a)    V = - x ( σ / 2ε₀)

c)  parallel to the flat sheet of paper

Explanation:

a) For this exercise we use the relationship between the electric field and the electric potential

          V = - ∫ E . dx        (1)

for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet

       Ф = ∫ E . dA = q_{int} /ε₀

the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product

          E A = q_{int} /ε₀

area let's use the concept of density

        σ = q_{int}/ A

       q_{int} = σ A

          E = σ /ε₀

as the leaf emits bonnet towards both sides, for only one side the field must be

          E = σ / 2ε₀

         we substitute in equation 1 and integrate

      V = - σ x / 2ε₀  

       V = - x ( σ / 2ε₀)

if the area of ​​the sheeta is 100 cm² = 10⁻² m²

      V = - x  (10⁻²/(2 8.85 10⁻¹²) = - x  ( 5.6 10⁻¹⁰)

       

      x = 1 cm     V = -1   V

      x = 2cm     V = -2   V

This value is relative to the loaded sheet if we combine our reference system the values ​​are inverted

       V ’= V (inf) - V

       x = 1 V = 5

       x = 2 V = 4

       x = 3 V = 3

   

These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.

 

In the attachment we can see a schematic representation of the equipotential surfaces

b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced

c) parallel to the flat sheet of paper

8 0
3 years ago
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