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monitta
4 years ago
6

What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.

a more recent study has reported an average temperature of 98.2°f. what is the difference between these averages, expressed in celsius degrees?
Physics
3 answers:
harkovskaia [24]4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

Igoryamba4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

givi [52]4 years ago
6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC OR

98.6*5ºC/9ºF = 54.80 98.25ºC/9ºF = 54.60 54.80 - 54.60 = .2ºC

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Which of the following is true for valence electrons?
Debora [2.8K]

Answer:

<u>Valence electrons are always located in the outer most energy level.</u>

Explanation:

Valence electrons are the ones that are involved in chemical bonds. In order to take part in a chemical bonding, the outermost/valence electron needs to be involved. Thus, the answer is <u>Valence electrons are always located in the outer most energy level.</u>

3 0
3 years ago
One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the o
max2010maxim [7]

To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

\text{Length of the string} = L = 4.32 m

\text{Frequency of the wave} = f = 75 Hz

\text{Time taken to reach the other end} = t = 0.5 s

Velocity of the wave,

V = \frac{L}{t}

V = \frac{4.32 m}{0.5s}

V = 8.64m/s

Wavelength of the wave,

\lambda = \frac{V}{f}

\lambda = \frac{8.64m/s}{75Hz}

\lambda = 0.1152m

\lambda = 11.52cm

Therefore the wavelength of the waves on the string is 11.53 cm

4 0
4 years ago
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3 years ago
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Ask Your Teacher A coil is formed by winding 260 turns of insulated 16 gauge copper wire (diameter = 1.3 mm) in a single layer o
Liula [17]

Answer:

The resistance is 2.98Ω

Explanation:

<u>Data</u>

diameter:  d=1.3mm, d=1.3*10^{-3} m

Radius:    r=\frac{d}{2} ,r=\frac{1.3*10^{-3} }{2} ,r=6.5*10^{-4} m

Resistivity constant:   p=1.65*10^{-8}Ωm

To find Area:

A=\pi r^{2} \\A=\frac{22}{7} *(6.5*10^{-4} )\\A=1.33*10^{-6} m^{2}

To find circumference:

l=260*(2\pi (0.15))\\l=245.04m

Now we can apply the Resistivity formula:

R=p\frac{l}{A}

R=(1.65*10^{-8} )Ωm*\frac{240.04m}{1.33*10^{-6}m^{2}  }

R=2.98Ω

4 0
4 years ago
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