Physics College

¿Cuál es el trabajo neto en J que se necesita para acelerar un auto de 1500 kg de 55 m/s a 65 m/s?

What is the net work in J required to accelerate a 1500 kg car from 55 m/s to 65 m/s?

Answers

Answer 1

The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J

How do i determine the net work done?

First, we shall obtain the initial kinetic energy. Details below:

Mass (m) = 1500 Kginitial velocity (u) = 55 m/sInitial kinetic energy (KE₁) =?

KE₁ = ½mu²

= ½ × 1500 × 55²

= 41250 J

Next, we shall final kinetic energy. Details below:

Mass (m) = 1500 KgFinal velocity (v) = 65 m/sFinal kinetic energy (KE₂) =?

KE₂ = ½mv²

= ½ × 1500 × 65²

= 3168750 J

Finally, we shall determine the net work done. Details below:

Initial kinetic energy (KE₁) = 41250 JFinal kinetic energy (KE₂) = 3168750 JNet work done (W) =?

W = KE₂ - KE₁

= 3168750 - 41250

= 3127500 J

Thus, the net work done is 3127500 J

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Related Questions

The temperature of a aluminum bar rises by 10.0°C when it absorbs 4.73 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of aluminum from these data. Answer is in kJ/kg · °C.

Answers

Answer:

Certainly! We can use the formula:

q = mcΔT

where q is the amount of heat absorbed, m is the mass of the aluminum bar, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.

Substituting the given values, we get:

4.73 kJ = (0.525 kg) x c x (10.0°C)

Solving for c, we get:

c = 0.901 kJ/kg · °C

Therefore, the specific heat of aluminum is 0.901 kJ/kg · °C.

Explanation:

A satellite weighing 5,400 kg is launched into orbit 3.6400 x 107 m above the center of the earth.

The mass of Earth is 6.0 × 1024 kg. The gravitational constant is 6.673 × 10–11 N•m2/kg2.

The gravitational force of Earth on the satellite is ___



Group of answer choices

9.1 x 10^4

1.6 x 10^3

2.1 x 10^6

Answers

Answer:

[tex]\tt F=1.63*10^3 N[/tex]

Explanation:

Gravitational force is defined as the force of attraction between two objects with mass. It is a fundamental force of nature, and it is what keeps us on the ground and what keeps the planets in orbit around the Sun.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

For the Question:

We can use the following formula to calculate the gravitational force between the Earth and the satellite:

[tex]\boxed{\tt F =\frac{ G * M * m }{ r^2}}[/tex]

Where:

F is the gravitational force

G is the gravitational constant[tex]\tt (6.673 * 10^{-11} Nm^2/kg^2)[/tex]

M is the mass of the Earth [tex]\tt (6.0 * 10^24 kg)[/tex]

m is the mass of the satellite[tex]\tt (5,400 kg)[/tex]

r is the distance between the satellite and the center of the Earth [tex]\tt (3.6400 * 10^7 m)[/tex]

Plugging in these values, we get the following:

[tex]\tt F = \frac{6.673 * 10^{-11} * 6.0 * 10^{24}* 5,400 }{ (3.6400 * 10^7 )^2}[/tex]

[tex]\tt F=1.63*10^3 N[/tex]

Therefore, answer is [tex]\tt F=1.63*10^3 N[/tex]

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