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2 years ago

Two thermometers are calibrated, one in degrees Celsius and the other in degrees Fahrenheit.

Physics

1 answer:

sdas [7]2 years ago

8 0

Answer:

The temperature is 233.15 K

Explanation:

Recall the formula to convert degree Celsius (C) into Fahrenheit (F):

$\frac{9}{5} C+32=F$

So if we want the value of degree C to be the same as the value of the degree F, we want the following: C = F

which replacing F with C on the right hand side of the equation above, allows us to solve for C:

$\frac{9}{5} C+32=F\\\frac{9}{5} C+32=C\\\frac{9}{5} C-\frac{5}{5} C =-32\\\frac{4}{5} C==32\\C= \frac{-32\,*\,5}{4} \\C=-40$

This means that -40°C = -40°F

And this temperature in Kelvin is:

-40°C + 273.15 = 233.15 K

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A ship sends out a 1200 Hz sound wave, which has a wavelength of 120 cm in the water. What would happen if that ship sent out a

OlgaM077 [116]

Answer:

the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

Explanation:

From wave,

v = λf................ Equation 1

Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: f = 1200 Hz, λ = 120 cm = 1.2 m

Substitute into equation 1

v = 1200(1.2)

v = 1440 m/s.

When the ship sent out a 600 Hz sound wave,

make λ the subject of formula in equation 1

λ = v/f............. Equation 2

Given: f = 600 Hz, v = 1440 m/s

Substitute into equation 2

λ = 1440/600

λ = 2.4 m or 240 cm.

When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm

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2 years ago

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

tatuchka [14]

Explanation:

It is given that,

Magnetic field, B = 0.5 T

Speed of the proton, v = 60 km/s = 60000 m/s

The helical path followed by the proton shown has a pitch of 5.0 mm, p = 0.005 m

We need to find the angle between the magnetic field and the velocity of the proton. The pitch of the helix is the product of parallel component of velocity and time period. Mathematically, it is given by :

$p=v_{||}\times T$

$p=v\ cos\theta\times \dfrac{2\pi m}{Bq}$

$cos\theta=\dfrac{pBq}{2\pi mv}$

$cos\theta=\dfrac{0.005\times 0.5\times 1.6\times 10^{-19}}{2\pi \times 1.67\times 10^{-27}\times 60000}$

$\theta=50.58^{\circ}$

So, the angle between the magnetic field and the velocity of the proton is 50.58 degrees. Hence, this is the required solution.

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2 years ago

A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n

Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

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

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

$\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m$

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

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

where for the dolphin:

v = 1531 m/s is the wave speed

$\lambda=1.50 cm=0.015 m$ is the wavelength

Substituting into the equation,

$f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz$

C. 13.6 m

Again, the wavelength is given by:

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

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

$\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m$

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

$\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m$

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

$\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m$

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

$\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m$

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

$f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz$

3 0

2 years ago

A floating leaf oscillates up and down two complete cycles in one second as a water wave passes by. The wave's wavelength is 10

postnew [5]

Answer:

C) 20 m/s

Explanation:

Wave: A wave is a disturbance that travels through a medium and transfers energy from one point to another, without causing any permanent displacement of the medium itself. Examples of wave are, water wave, sound wave, light rays, radio waves. etc.

The velocity of a moving wave is

v = λf ............................ Equation 1

Where v = speed of the wave, λ = wave length, f = frequency of the wave.

Given: f = 2 Hz (two complete cycles in one seconds), λ = 10 meters

Substituting these values into equation 1

v = 2×10

v = 20 m/s.

Thus the speed of the wave = 20 m/s

The right option is C) 20 m/s

7 0

2 years ago

A simple pendulum 0.64m long has a period of 1.2seconds. Calculate the period of a similar pendulum 0.36m long in the same locat

weqwewe [10]

The period of the second pendulum is 0.9 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity at the location of the pendulum

For the first pendulum, we have

L = 0.64 m

T = 1.2 s

Therefore we can find the value of g at that location:

$g=(\frac{2\pi}{T})^2 L=(\frac{2\pi}{1.2})^2 (0.64)=17.5 m/s^2$

Now we can find the period of the second pendulum at the same location, which is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where we have

L = 0.36 m (length of the second pendulum)

$g=17.5 m/s^2$

Substituting,

$T=2\pi \sqrt{\frac{0.36}{17.5}}=0.9 s$

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8 0

2 years ago