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2 years ago

How is energy that is stored in food released?

Physics

2 answers:

kumpel [21]2 years ago

7 0

Answer:

i think its A

Explanation:

Triss [41]2 years ago

3 0

Answer:

Explanation:

The chemical energy stored in food is released by cells through the process of respiration. (cellular respiration)

This stored energy is released whenever these chemical bonds are broken in metabolic processes such as cellular respiration. Cellular respiration is the process by which the chemical energy of “food” molecules is released and partially captured in the form of ATP.

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Takumi works in his yard for 45 minutes each Saturday. He works in the morning, and he wears sunscreen and a hat each time he wo

MrRa [10]

Explanation :

Takumi wears sunscreen and a hat each time he works in the yard. This is to protect himself with the strong radiation coming from the sun. UV rays that are coming from the sun are the main cause of skin cancer.

Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.

So, the correct option is (b) "the severity of stochastic effects, such as cancer".

7 0

3 years ago

Read 2 more answers

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

3 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

2 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

2 years ago

What is a quantum field?

liubo4ka [24]

It is the mathematical and conceptual framework for contemporary Elementary practical physicas. if that make sense to you.

8 0

3 years ago

Read 2 more answers