Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook

Anna71 [15]

Answer:

The mass flow rate is 2.37*10^-4kg/s

The exit velocity is 34.3m/s

The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW

7 0

2 years ago

Define work What is the unit for work? For Physics, thanks.

Minchanka [31]

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement

Explanation:

5 0

3 years ago

Energy Transformation

Serjik [45]

Answer:

K =400000 J

Explanation:

Kinetic Energy

Is the energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The car has a mass of m=2000 Kg and travels at v=20 m/s. Calculating the kinetic energy:

$\displaystyle K=\frac{1}{2}2000*20^2$

Calculating:

K =400000 J

3 0

3 years ago

One of the foci for the moon's orbit would be the

suter [353]

The question is oversimplified, and pretty sloppy.

Relative to the Earth . . .
The Moon is in an elliptical orbit around us, with a period of
27.32... days, and with the Earth at one focus of the ellipse.

Relative to the Sun . . .
The Moon is in an elliptical orbit around the Sun, with a period
of 365.24... days, and with the Sun at one focus of the ellipse,
and the Moon itself makes little dimples or squiggles in its orbit
on account of the gravitational influence of the nearby Earth.

I'm sorry if that seems complicated. You know that motion is
always relative to something, and the solar system is not simple.

5 0

3 years ago

Read 2 more answers

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago