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2 years ago

How much energy is required to raise the temperature of 50.0 grams of water 10.0 degree C? (Explain yourself answer in joules!)

Physics

1 answer:

maria [59]2 years ago

4 0

The amount of energy needed is 2093 J

Explanation:

The amount of energy needed to increase the temperature of a substance by $\Delta T$ is given by the equation

$Q=mC\Delta T$

where

m is the mass of the substance

C is its specific heat capacity

$\Delta T$ is the increase in temperature

For the water in this problem, we have

m = 50.0 g = 0.050 kg

$C=4186 J/g^{\circ}C$ (specific heat capacity of water)

$\Delta T=10.0^{\circ}C$

Therefore, the amount of energy needed is

$Q=(0.050)(4186)(10)=2093 J$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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A wooden block has a mass of 562 g and a volume of 72 cm3. what is the density

Gnom [1K]

Answer:

7 $\frac{29}{36}$ g/cm³

Explanation:

The formula for density is $\frac{mass}{volume}$ . Let's apply the formula to the question:

$\frac{562}{72}$ = 7 $\frac{29}{36}$ g/cm³

7 0

2 years ago

Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the sp

Alekssandra [29.7K]

Answer:

2.92682 m

$1.5\times 10^{18}\ Hz$

250000000000 Hz

$2.88462\times 10^{-8}\ m$

0.21126 m

0.12244 m

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength

f = Frequency

Wavelength is given by

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m$

The wavelength is 2.92682 m

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz$

The frequency is $1.5\times 10^{18}\ Hz$

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{1.2\times 10^{-3}}\\\Rightarrow f=250000000000\ Hz$

The frequency is 250000000000 Hz

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.04\times 10^{16}}\\\Rightarrow \lambda=2.88462\times 10^{-8}\ m$

The wavelength is $2.88462\times 10^{-8}\ m$

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m$

The wavelength is 0.21126 m

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{2.45\times 10^9}\\\Rightarrow \lambda=0.12244\ m$

The wavelength is 0.12244 m

8 0

2 years ago

The car traveled part of the route at an average speed of 60 km / h. then it traveled at 130 km / h, and the last section travel

frutty [35]

Answer:

Explanation:

200/2.5

=80

7 0

2 years ago

Which atoms have the lowest electronegativity?

jek_recluse [69]

Answer: Option A) Atoms to the left on the periodic table

Explanation:

The periodic table has metallic elements on the left side, semi-metal or metalloids in the middle, and nonmetallic elements of the right side.

Since, electronegativity of an atom tells of its power to attract electrons. Metals give off electrons rather than attract electrons like non-metals; thus, metals have the lowest electronegativity (with values less than 2.4 while nonmetal have high electronegativity values above 2.7

Infact, the element with the highest electronegativity is a non-metal (fluorine) while the element with the lowest electronegativity is a metal (potassium)

Thus, metallic atoms located on the left side of the periodic table have the lowest electronegativity

6 0

2 years ago

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In a section of horizontal pipe with a diameter of 3.00 cm the pressure is 5.21 kPa and water is flowing with a speed of 1.50 m/

kvasek [131]

Answer:

The pressure at the narrower region reduces to 4.00 kPa

Explanation:

According to Bernoulli's theorem for an ideal fluid we have we have

$\frac{P}{\gamma _{w}}+\frac{v^{2}}{2g}+z=constant$

Applying this between the two sections under consideration we obtain

$\frac{P_{1}}{\gamma _{w}}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{v_{2}^{2}}{2g}+z_{2}$

Now by equation of contuinity we have

$A_{1}v_{1}=A_{2}v_{2}$

$\therefore v_{2}=\frac{A_{1}v_{1}}{A_{2}}\\\\\therefore v_{2}=\frac{1.5\times 0.25\times \pi \times (3)^{2}}{0.25\times \pi \times (2.50)^{2} }\\\\\therefore v_{2}=2.16m/s$

Applying values in the bernoulli's equation and noting that $z_{1}=z_{2}$ we get

$\frac{P_{2}}{\gamma _{w}}=\frac{5.21\times 10^{3}}{1000\times 9.81}+\frac{1.50^{2}}{2g}-\frac{2.16^{2}}{2g}\\\\\therefore \frac{P_{2}}{\gamma _{w}}=0.40797\\\\\therefore P_{2}=4.00kPa$

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3 years ago

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