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shepuryov [24]
2 years ago
11

Part a a calorimeter consists of an aluminum cup inside of an insulated container. The cup is weighed on a top-loading balance a

nd is found to have a mass of 31. 91 g. A reaction is conducted in the calorimeter, raising the temperature from 21. 2 c to 26. 1 c what is the change in heat q for the aluminum cup in units of j? aluminum has a specific heat of 0. 903 j 1 ? 1 write your answer to the correct number of significant figures.
Physics
1 answer:
vovikov84 [41]2 years ago
5 0

The change in heat q for the aluminum cup is 141 J.  

<h3>What is Specific heat?</h3>

A substance's heat capacity is the amount of heat needed to increase its overall temperature by one degree. The heat capacity is known as the specific heat capacity or the specific heat if the substance's mass is unity.

change in heat -

The amount of heat Q that is transported to create a temperature shift depends on the substance and phase involved, the mass of the system, and the magnitude of the temperature change. (A) The temperature change and the amount of heat transported are directly inversely related.

The answer's sign dig is determined by a calculation's lowest sign digits.

The solution will need 3 digits because the temperatures are provided with 3 significant digits and the specified heat has 3 significant digits (the mass has 4 significant digits).

31.91 g   * 0.903 J/g-C  *  (26.1 - 21.2 C) = 141 J

to learn more about Specific Heat go to - brainly.com/question/21406849

#SPJ4

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Answer:

Part 1

20 N

Part 2

0.4 m/s²

Part 3

4 m/s

Explanation:

The force which pulls the sled right = 50 N

The friction force exterted towards left by the snow = -30 N

The mass of the sled = 50 kg

Part 1

The sum of the forces on the sled, F = 50 N + (-30) N = 20 N

Part 2

The acceleration of the sled is given as follows;

F = m·a

Where;

m = The mass of the sled

a = The accelertion

a = F/m

∴ a = (20 N)/(50 kg) = 0.4 m/s²

The acceleration of the sled, a = 0.4 m/s²

Part 3

The initial velocity of the sled, u = 2 m/s

The kinematic equation of motion to determine the speed of the sled is v = u + a·t

The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;

v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.

7 0
3 years ago
An athlete runs at a constant velocity of 5.2 m/s. What is the velocity of the athlete relative to the ground?
dimulka [17.4K]

The relative velocity of the athlete relative to the ground is 5.2 m/s

The given parameters;

constant velocity of the athlete, V = 5.2 m/s

let the velocity of the ground = Vg = 0

The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.

The athlete is the moving object in this question while the ground is stationary.

The relative velocity of the athlete relative to the ground is calculated as follows;

V/V_g = V - V_g  = 5.2 - 0 = 5.2  \ m/s

Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s

Learn more here: brainly.com/question/24430414

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Coders play an important role in
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it is a.health record documentation

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4 years ago
An Aichi D3A bomber, with mass of 3600 kg, departs from its aircraft carrier with a velocity of 85 m/s due east. What is the pla
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7 0
3 years ago
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

1000\times\dfrac{0.10400}{0.02}=\rho_{m}

\rho_{m}=5200\ kg/m^3

Hence, The density of the metal is 5200 kg/m³.

6 0
4 years ago
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