Ask question

Ask question

All categories

3 years ago

8

A 0.50-g sample of vegetable oil is placed in a calorimeter. when the sample is burned, 18.9 kj is given off. what is the energy

value (kcal/g) for the oil?

1 answer:

just olya [345]3 years ago

8 0

First of all, we need to convert the energy produced from kilojoules to calories. 1 kJ corresponds to 239.1 cal, therefore:

$E=18.9 kJ \cdot 239.1 =4519.4 cal$

And so now we can calculate the energy value of the oil:

$E_v = \frac{E}{m}=\frac{4519.4 cal}{0.50 g}=9038.8 cal/g$

You might be interested in

What are the characters associated with light as a wave

Liono4ka [1.6K]

Answer:

Interference of light

Diffraction of light

Polarization of light

Reflection of light

all show the wave nature of light.

6 0

3 years ago

Light from a 560 nm monochromatic source is incident upon the surface of fused quartz (n=1. 56) at an angle of 60°. what is the

ankoles [38]

The angle of reflection is "60°".

Here we apply the law of the concept of reflection then we get the final answer easily.

The angle of incident = angle of reflection

Then, the Angle of the incident =60°

What is reflection?

Reflection is the phenomenon of light rays returning to the source after striking an obstruction.
It resembles the way a ball bounces when we toss it on a hard surface.
Some of the light rays that strike an item are reflected, some of them travel through it, and the remainder are absorbed by the object.
The given values are:Light from a monochromatic source,= 560 nm
The angle of incidence,= 60°
The surface of fused quartz (n),= 1.56
When a light ray does exist on a flat surface, the law or idea of reflection should apply since it includes both the reflected and "normal" light rays at the mirror surface.
According to the above law,Angle of incident = angle of reflection
Then, Angle of incident =60°.

To learn more about reflection visit: brainly.com/question/15487308

#SPJ4

7 0

2 years ago

A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant

GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude <em>w</em> = 43.2 N)

• the normal force, pointing up (mag. <em>n</em>)

• the applied force, pointing the direction in which the box is sliding (mag. <em>p</em> = 6.30 N, with <em>p</em> for "pull")

• the frictional force, pointing oppoiste the applied force (mag. <em>f</em> )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

<em>n</em> + (-<em>w</em>) = 0

and

<em>p</em> + (-<em>f</em> ) = 0

So then the forces have magnitudes

<em>w</em> = 43.2 N

<em>n</em> = <em>w</em> = 43.2 N

<em>p</em> = 6.30 N

<em>f</em> = <em>p</em> = 6.30 N

5 0

3 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago

How much work is required to lift a 2 kg mass to a height of 10 meters?

scoray [572]

Use the equation potential energy =m*g*h
m-mass,
g-gravitational acceleration,
h-height
Potential energy = 2*10*10
=200
$kg {m}^{2} {s}^{ - 2}$
This is the unit of the energy

6 0

3 years ago