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4 years ago

11

The initial temperature of a bomb calorimeter is 28.50°

2 answers:

solniwko [45]4 years ago

8 0

The answer is 5,170 J

DaniilM [7]4 years ago

3 0

Answer:

The answer is 5174400 J.

Explanation:

Here you have to keep in mind the equation to determine the amount of heat (Q) that one body of mass <em>m </em>can take. The equation is : [tex] Q=mc\Delta T [\tex]

where <em>m</em> is the mass, <em>c</em> is the specific heat and [tex] \Delta T [\tex] the temperature variation. This way: Q=1400kg*3.52J/gºC*(28.5-27.45) ºC=1400000 g*3.52 J/gºC * (28.5-27.45)ºC=5174400J=5174.4 kJ=5.17 MJ.

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Which until is commonly used to measure density

Ivenika [448]

Separate mass and volume

6 0

4 years ago

Read 2 more answers

Un hombre parado en el techo de un edificio, tira una bola verticalmente hacia arriba con una velocidad de 12m/s, la bola llega

zmey [24]

Respuesta:

24m

Explicación:

Según la ecuación de movimiento

v = u + en

Dado

Velocidad final v = 12 m / s

velocidad inicial u = 0 m / s

tiempo t = 4s

Sustituir

12 = 0 + 4a

a = 12/4

a = 3 m / s²

Lo siguiente es obtener la distancia;

S = ut + 1 / 2at²

S = 0 (4) + 1/2 (3) (4) ²

S = 3 * 16/2

S = 48/2

S = 24 m

Por lo tanto, la distancia requerida es de 24 m.

4 0

3 years ago

Calculate the mass of air of density 1.2kgm-3 in a room of dimensions 8m by 6m by 4m.

77julia77 [94]

Answer:

230.4kg

Explanation:

volume of the room = l× b×h

volume= 8×6×4

volume=192m3

density= mass/volume

hence mass= density × volume

mass= 1.2kgm-3 × 192m3

mass= 230.4kg

5 0

2 years ago

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by

Artemon [7]

Answer:

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

$E=I^2Rt$ . ....(1)

As,

$I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A$

As per Ohm's law, V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega$

Using formula (1).

$E=0.835^2\times 14.37\times 2\\\\=20.03\ J$

So, the energy delivered to the lightbulb filament is 20.03 J.

6 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago