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3 years ago

Where have you heard the word calorie before what do you think a calorie is

Physics

1 answer:

Ainat [17]3 years ago

6 0

Answer:

Explanation:

Calorie is a common term used to describe the amount of energy that can be derived from food products.

We quantify foods based on the calories of energy they possess. A high calorific food will yield more energy to the body and is often desired for intense physical activities.

Calorie is defined as the amount of heat energy needed to raise the temperature of 1g of a substance by 1°C.

Food calories slightly differs in definition and they imply the amount of heat needed to raise the temperature of 1kg of water by 1°C.

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Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

e-lub [12.9K]

Answer:

$f = 421.8 Hz$

Explanation:

When she moved a distance of 1 m from mid point she observe first destructive interference due to two speakers

so we can say that path difference of sound due to two speakers will be equal to half of the wavelength

so path difference is given as

$\Delta L = {3.5^2 + 12^2}^{0.5} - {1.5^2 + 12^2}^{0.5}$

so it will be

$\Delta L = 12.5 - 12.093$

$\Delta L = 0.4066$

now we know that

$\frac{\lambda}{2} = 0.4066$

$\lambda = 0.813$

now frequency of sound is given as

$f = \frac{v}{\lambda}$

$f = \frac{343}{0.813}$

$f = 421.8 Hz$

4 0

3 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)