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11 months ago

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

Physics

1 answer:

iris [78.8K]11 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

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Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

3 years ago

Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan

Sonja [21]

<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

$R_{A}$ = Resistance in bulb A = 3.0Ω

$R_{B}$ = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance ( $R_{X}$ ) is given by

$\frac{1}{R_{X}}$ = $\frac{1}{R_{A} }$ + $\frac{1}{R_{B} }$ ---------------(i)

Substitute the values of $R_{A}$ and $R_{B}$ into equation (i)

=> $\frac{1}{R_{X}}$ = $\frac{1}{3.0}$ + $\frac{1}{2.5}$

Solve for $R_{X}$

$R_{X}$ = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = $R_{X}$ = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4 = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

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