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Luba_88 [7]
2 years ago
6

A computer simulation is an example of what?

Physics
1 answer:
Lelechka [254]2 years ago
8 0
<h3>A computer simulation is an example of a <u>Model</u>.</h3>

A model can be a virtual OR physical representation of an object, movement, etc. A model is used to have a visual view of an item.

Hope this helps and please mark brainliest!

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Why is Florida more vulnerable to the effects of climate change? ​
AURORKA [14]

Answer:

Along the Atlantic and Gulf Coasts of Florida, the land surface is also sinking. If the oceans and atmosphere continue to warm, sea level along the Florida coast is likely to rise one to four feet in the next century. Rising sea level submerges wetlands and dry land, erodes beaches, and exacerbates coastal flooding.

Explanation:

6 0
3 years ago
A light bulb whose resistance is 240 ohms is connected to a 120 voltage source. What is the current through the bulb?
Sedaia [141]

Answer:

0.5A

Explanation:

Using R = \frac{V}{I},

R is the resistance (in Ohms)

V is the voltage (in V)

I is the current (in A)

240=\frac{120}{I}

I = 0.5A

7 0
2 years ago
If 2.40 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pre
creativ13 [48]

Answer:

-112.876J

Explanation:

In order to solve this question, we would need to incorporate Stoichiometry, which involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Here's a balanced equation for the reaction:

16KNO_3(s) + 24C(s) + S_8(s)    \to 24CO_2(g) + 8N_2(g) + 8K_2S(s)

Let us define P - V work as;

w_{pv} = - P_{external}  \triangle Volume

where  \triangle (Volume) = (V_{final} - V_{initial})

External pressure is given as  1.00atm , therefore the work solely depends on the change in volume and since the reactants are solids, none of the reactants contribute to the volume. Hence,  V_i = 0.

To find the volume of the products, we need to first find the amount of moles of the product made from  2.40_gKNO_3, using the molar mass of  KNO_3  which is 101.1032 g/mol  

2.40_gKNO_3 . {\frac{1molKNO_3}{101.1032_g}} = 0.0237molKNO_3

Now let us convert moles of  KNO_3  into moles of CO_2 and N_2  using the stoichiometric ratios from our balanced equation of the reaction.

0.0237molKNO_3 . {\frac{24molCO_2}{16molKNO_3}} = 0.0356molCO_2

0.0237molKNO_3 . {\frac{8molN_2}{16molKNO_3}} = 0.01185molN_2

K_2S is not factored into the volume calculation because it is a solid.

Now let us also convert the moles of  CO_2  and  N_2 into grams using their respective molar masses.

0.0356molCO_2 . {\frac{44.01_g}{1molCO_2}} = 1.567_gCO_2

0.01185molN_2 . {\frac{28.014_g}{1molN_2}} = 0.332_gN_2

We will now proceed to convert grams into volume using the density values provided.

1.567_gCO_2 . {\frac{1L}{1.830_g}} = 0.856LCO_2

0.332_gN_2 . {\frac{1L}{1.165_g}} = 0.285LN_2

Summing up the two volumes, we get the final volume

0.856L + 0.258L = 1.114L = V_f

Plugging everything into the w_{pv} equation, we get:

w_{pv} = -1atm(1.114L - 0L) = -1.114L.atm

Finally, let us convert L.atm into joules using the conversion rate of;

1L.atm = 101.325J\\-1.114L.atm. {\frac{101.325J}{1L.atm}} = -112.876J

7 0
3 years ago
A car starting from rest accelerated at a rate of 0 .5 m/s up to 2 km what would be the final velocity of the car and how much t
Julli [10]

From equation of motion v^2 = u^2 +2aS

Hence, the final velocity is 40 m/s.

<em><u>hope </u></em><em><u>it's </u></em><em><u>help </u></em><em><u>you</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u>! </u></em>

<em><u>#</u></em><em><u>rishu</u></em>

4 0
3 years ago
Read 2 more answers
Point masses m1 m2 are placed at opposite ends
tiny-mole [99]

(a) x = \frac{m_2L}{m_1+m_2}

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at \frac{m_2L}{m_1 + m_2}

7 0
3 years ago
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