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3 years ago

Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.

Physics

1 answer:

Viktor [21]3 years ago

6 0

Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

$K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J$

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Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

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3 years ago

Behavior that is harmful to one's self is? 10 pts

Sophie [7]

The answer is distressing

8 0

3 years ago

Read 2 more answers

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential

Citrus2011 [14]

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick = 7.9kg

Height of building = 22m

Unknown:

Potential energy of the brick = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

Insert the given parameters and solve;

P.E = 7.9 x 9.8 x 22 = 1703.24J

6 0

3 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

3 0

3 years ago