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Arlecino [84]
4 years ago
10

Is the temperature of a normal healthy old man is less than that of a normal healthy young man?

Physics
1 answer:
Inessa [10]4 years ago
8 0

Answer:

The correct answer is no.

Explanation:

The temperature of a person changes depending on their age, the physical activity they are doing and until the time of day.

There is an established average temperature that is 98.6 ° F (37 ° C) as a healthy temperature.

If the temperature exceeds 100.4 ° F (38 ° C) it means that you may be ill or have an infection.

The difference in temperature by age is very minimal, but it is still a difference. Let's see:

  • Babies and children: from 97.9 ° F (36.6 ° C) to 99 ° F (37.2 ° C).
  • Adults: from 97 ° F (36.1 ° C) to 99 ° F (37.2 ° C).
  • Adults over age 65: 98.6 ° F (36.2 ° C).

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4 years ago
A wave is represented by the equation y = 0.20 sin [ 0.4π (x – 60t)], where x and y are in cm and t is in s. The velocity of the
Leya [2.2K]

Answer:

v = 60 m/s

Explanation:

It is given that,

A wave is represented by the equation :

y=0.2\sin [0.4\pi (x-60)t]

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The general equation of a wave is given by :

y=A\sin (kx-\omega t) ....(1)

Equation (1) can be written as :

y=0.2\sin (0.4\pi x-24\pi t) ...(2)

If we compare equation (1) and (2) we get :

k=0.4\pi

\omega=24\pi

The velocity of a wave is given by :

v=\dfrac{\omega}{k}\\\\v=\dfrac{24\pi}{0.4\pi}\\\\v=60\ m/s

So, the velocity of the wave is 60 m/s.

6 0
3 years ago
When large amplitude sound vibrates the ear significantly, a _______ results.
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4 0
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You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea
KonstantinChe [14]

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant  = ?

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a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

k=\frac{2.303}{90.3min}\log\frac{400}{100}

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Now we have to calculate the half-life of substance, we use the formula :

k=\frac{0.693}{t_{1/2}}

1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=45min

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