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lara31 [8.8K]
10 months ago
15

The student used two different types of thermometer to measure the temperature

Physics
1 answer:
iris [78.8K]10 months ago
5 0

0.02 °C  is the resolution of thermometer B while measuring temperature.

Thermometer A= 87.6

Thermometer B=87.4°C

resolution= 87.6- 87.4°C = 0.02 °C

Heat or cold are measured in terms of temperature. It can be defined in terms of a variety of arbitrary scales and shows how heat energy naturally moves from a hotter to a cooler substance (one at a lower temperature). Even if a match burns at a considerably greater temperature than an iceberg, an iceberg has a much higher overall heat energy. The energy of a thermodynamic system differs from temperature.

Temperature, along with pressure, density, and other related qualities, is referred to as an intense property—one that is independent of the amount of material being addressed—in contrast to broad characteristics like mass or volume.

To know more about  temperatures visit : brainly.com/question/14161300

#SPJ1

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In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

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\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

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3 years ago
3.
abruzzese [7]

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3 years ago
A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin
Mashcka [7]

Answer:

r_{cm}=[12.73,12.73]cm

Explanation:

The general equation to calculate the center of mass is:

r_{cm}=1/M*\int\limits {r} \, dm

Any differential of mass can be calculated as:

dm = \lambda*a*d\theta  Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}

So, the differential of mass is:

dm = \frac{2M}{a\pi}*a*d\theta

dm = \frac{2M}{\pi}*d\theta

Now we proceed to calculate X and Y coordinates of the center of mass separately:

X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta

Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta

Solving both integrals, we get:

X_{cm}=2*a/\pi=12.73cm

Y_{cm}=2*a/\pi=12.73cm

Therefore, the position of the center of mass is:

r_{cm}=[12.73,12.73]cm

5 0
2 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

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magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

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Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

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