¿Existe relación entre la rapidez para realizar trabajo y la velocidad del cuerpo?

An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy a

Shtirlitz [24]

Explanation:

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

$\bold{↬{ }}$ ${\bold{\sf{\underline{Given \: that}}}}$

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

$\bold{↬{ }}$ ${\bold{\sf{\underline{To \: find}}}}$

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

${\bold{\sf{\underline{Solution}}}}$

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

${\bold{\sf{\underline{Using \: concepts}}}}$

✰ Potential energy formula.

${\bold{\sf{\underline{Using \: formula}}}}$

✰ Potential energy = mgh

${\bold{\sf{\underline{We \: also \: write \: these \: as}}}}$

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

${\bold{\sf{\underline{Full \: solution}}}}$

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0

3 years ago

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Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n

pickupchik [31]

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0

3 years ago

Ryan runs 50 m north and then 50 m east. What is his displacement?

san4es73 [151]

Answer:

Explanation:

Distance and direction of an object's change in position from a starting point. Displacement. 3. Jermaine runs exactly 2 laps around a 400 meter track. What is the displacement? 0 ... David walks 3 km north, and then turns east and walks 4 km. ... A person walks 50 meters directly north, stops, and then travels 32 meters

3 0

2 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

Define compound and list two examples.

Liono4ka [1.6K]

A compound is a substance that consists of two or more elements, which is chemically combined, meaning that it could only be split by chemical means.

Examples:-

- NaCl is salt, chemically combined of Sodium and Chloride.

- H2O is water, chemically combined of Hydrogen and oxygen.

8 0

3 years ago

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