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2 years ago

Another word for mechanical weathering is

Physics

1 answer:

aliya0001 [1]2 years ago

6 0

1.disintegration
2.soil
3.oxidation
4.freezing and thawing

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How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•

Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

$Q = 12*4.184*15.57$

Q = 781.74 Joules

7 0

2 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

2 years ago