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Korolek [52]
3 years ago
6

How much does 100g grams of platinum cost

Physics
1 answer:
Bumek [7]3 years ago
7 0
Today (April 28), platinum closed at $1,012 per troy ounce,
and gold closed at $1,243 the ounce.

1 troy ounce = 31.1 grams.

So 100g of platinum would be worth  (100/31.1) x $1,012  =  $3,254

and 100g of gold would be worth  (100/31.1) x $1,243  =  $$3,997 (rounded). 

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A cook preparing a meal for a group of people is an example of ______
Sati [7]

Answer:

Option B is correct.

Explanation:

Potential energy or chemical potential energy is used to Cook food which is then converted into thermal energy. The type of energy used also depends upon the type of cooking appliances used.

For e.g. stove convert potential energy to thermal energy.

7 0
3 years ago
The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t
REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

                          = -177.275J

   

7 0
2 years ago
Read 2 more answers
A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
2 years ago
Una onda tiene una frecuencia de 350 Hz. ¿Cuál es su período?​
Alika [10]

Answer:0.00285714285 seconds

Explanation:

period=1 ➗ frequency

Period=1 ➗ 350

Period=0.00285714285 seconds

7 0
3 years ago
A car veers off course and runs straight into a brick wall. This is an example
ValentinkaMS [17]
Short time large force
4 0
3 years ago
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