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2 years ago

Hi, I was wondering what is the use of a temperature sensor since the sensor wouldn't be fully immersed in the hydrogen.

Physics

1 answer:

777dan777 [17]2 years ago

6 0

Answer:

Within our homes, temperature sensors are used in many electrical appliances, from our refrigerators and freezers to help regulate and maintain cold temperatures as well as within stoves and ovens to ensure that they heat to the required levels for cooking, air confectioners/heaters.

Explanation:

A temperature sensor is a device used to measure temperature. This can be air temperature, liquid temperature or the temperature of solid matter. There are different types of temperature sensors available and they each use different technologies and principles to take the temperature measurement.

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A CAR ACCElerates FROM REST To

Ilia_Sergeevich [38]

Answer:

Explanation:

<u>Power Formula</u>

Power = Force × Velocity

<u>Calculating Force</u>

F = ma
F = m(v - u / t) [From the equation v = u + at]
F = 2,000 (50/7.5)
F = 2,000 (20/3)
F = 40,000/3
F = 13333.3 N

<u>Solving for Power</u>

P = 13333.3 × 50
P = 666666.6 W
P = <u>666.6 kW</u>

5 0

1 year ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

2 years ago

Read 2 more answers

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0

3 years ago

Which amusement park ride utilizes periodic motion?

hjlf

Periodic motion implies that the ride oscillates back and forth. The period is the time it takes to complete one cycle. Therefore it would be the swing.

6 0

2 years ago

How do you calculate relative velocity?

nekit [7.7K]

Answer:

<h3>Vab=Va-Vb</h3><h3>Vab=Vb-Va</h3><h3>Explanation:</h3>

Vab is velocity of a relative to b

Vba is velocity of b relative to a

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3 years ago