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3 years ago

The power of the kettle was 2.6 kW

Physics

1 answer:

faltersainse [42]3 years ago

7 0

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

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An Na ion has a charge of +1 and an O ion has a charge of –2. How many Na ions does one O ion need to balance charges

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Because Na⁺ ion lacks an electron and O²⁻ has two extra electrons extra, to balance the charge, we need 2 Na⁺ ions.

All ions, atoms and molecules want to get to the minimum energy state, and that state is when the ion, atom, or molecule is neutral, that's why all of them want to balance their charges.

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3 years ago

Read 2 more answers

An aluminum wire is 5.87 m long and has a diameter of 6.07 mm.

antoniya [11.8K]

Answer: 5.72 x 10-3Ω

Explanation:

Hi, to answer this question, first we have to calculate the cross sectional area of the cable:

Diameter (D)=6.07 mm

Since: 1000mm = 1m

6.07mm/ 1000mm/m = 0.00607 meters

Area of a circle : π (d/2)^2

A = π (0.00607/2)^2= 0.000028937 m2

Resistance formula:

Resistance (R) = P(resistivity) L (length)÷A (cross sectional area )

Replacing with the values given:

R = (2.82x10-8 x 5.87) / 0.000028937

R = 5.72 x 10-3Ω

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

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3 years ago

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

mixer [17]

Answer:

The kinetic energy of the ejected electrons increases.

Explanation:

As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.

Using conservation of energy principle we have

$E_{incident}=h\nu +\frac{1}{2}mv^{2}$

If we increase the intensity of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since $h\times \nu$ remains constant.