Someone help me please

How much heat is needed to raise the temperature of an empty 2.0 x 101 kg vat made of

scZoUnD [109]

Answer:

Q = 7272 Kilojoules.

Explanation:

<u>Given the following data;</u>

Mass = 2.0*101kg = 202kg

Initial temperature, T1 = 10°C

Final temperature, T2 = 90°C

We know that the specific heat capacity of iron = 450J/kg°C

*To find the quantity of heat*

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 = 90 - 10

dt = 80°C

Substituting the values into the equation, we have;

$Q = 202*450*80$

Q = 7272KJ or 7272000 Joules.

6 0

3 years ago

The sum of all the forces acting on an <br> object is zero.

marysya [2.9K]

Answer:

Newton's first law states that when the vector sum of all forces acting on an object (the net force) is zero, the object is in equilibrium. If the object is initially at rest, it remains at rest. If it is initially in motion, it continues to move with constant velocity.

Explanation:

3 0

2 years ago

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Please can someone help?

vodomira [7]

5207 because I said so

5 0

3 years ago

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

$Q=\frac{\lambda _{0}}{2L}$

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

4 0

3 years ago

Subduction zones form when an oceanic plate collides with another oceanic plate or continental plate. The continental crust is l

Fed [463]

The answer is; metamorphic

Occurs when igneous and sedimentary rocks are subjected to extreme temperatures and pressure such that their chemical composition changes. Examples of metamorphic rocks are; marble (that metamorphoses from limestone rock), quartzite (from sandstone). Other types of metamorphic rock include gneiss and schist.

5 0

3 years ago

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