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3 years ago

Is 5g greater than 20.43 mg

Physics

2 answers:

tatyana61 [14]3 years ago

7 0

Answer

Yes.

Explanation:

20.43mg is equal to 0.02043g.

worty [1.4K]3 years ago

4 0

yes that's true 6g is larger

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Para que exista una fuerza se necesita cuando menos…

kozerog [31]

Answer:

Explanation:

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3 years ago

A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s

Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

$Kinetic Energy, KE = \frac{1}{2} mv^2$

Therefore, $KE = \frac{62 X 5.48 X 5.48}{2}$

KE = 930.94 Joules

Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0

3 years ago

Which two events are part of the rock cycle?

kozerog [31]

Answer:

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2 years ago

Suppose that a uniform electric field exists in a certain region of space. Now consider a mathematical plane surface of area A.

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Answer: The normal of the plane must be parallel to the electric field vector.

Explanation:

the normal to the surface is defined as a versor that is perpendicular to the plane.

Now, if the angle between this normal and the line of the field is θ, we have that the flux can be written as:

Φ = E*A*cos(θ)

Where E is the field, A is the area and θ is the angle already defined.

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7 0

2 years ago

Two boxes m1 and m2 are connected by a lightweight cord and sitting on a frictionless surface. A horizontal force of 66 N pulls

Flura [38]

Answer:

m₂ = 18 kg

Explanation:

Conceptual analysis

We apply Newton's second law on masses m₁ and m₂

∑F = m × a Formula (1)

∑F: algebraic sum of forces in Newton (N)

m: mass in kg

a: acceleration in m/s²

T: cord tension in Newton (N)

Known data

m₁ = 12 kg

F = 66 N

a₁ = a₂ = 2.2 m/s²

Problem development

We can see in the attached figure the free body diagram of the masses and we replace the data in formula (1), considering that positive forces are the ones that go in the same direction of acceleration::

Mass 1 (m₁) free body diagram:

∑Fₓ = m₁ × a₁

66 - T = 12 × 2.2

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66 - 26.4 = T

T = 39.6 N

Mass 2 (m₂) free body diagram:

∑Fₓ = m₂ × a₂

T = m₂ × 2.2

39.6 = m₂ × 2.2

m₂ = 39.6 ÷ 2.2

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6 0

3 years ago