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katrin2010 [14]
3 years ago
10

What happens to your kinectic energy when your gravitational potential energy decreases?

Physics
1 answer:
leonid [27]3 years ago
3 0

Answer:

As r decreases, we lose gravitational potential energy - in other words, U G U_G UG​U, start subscript, G, end subscript becomes more negative. Because energy is conserved, the velocity must increase, resulting in an increase in kinetic energy.

Explanation:

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LuckyWell [14K]

Answer:d

Explanation:

7 0
3 years ago
Read 2 more answers
5. A man has a weight of 100 Newtons. How much work is done if he climbs 4 meters up a ladder? Plug numbers under the equation.
Virty [35]

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 100 × 4

We have the final answer as

<h3>400 J</h3>

Hope this helps you

5 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 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
What the kinetic Energy of 80kg<br> object this is Moving with a Speed<br> of 23 m/s
Katyanochek1 [597]

Explanation:

Kinetic Energy Formula = ½mv²

  • ½×800kg×(23m/s)²
  • 400kg×529m²/s²
  • 211600 kg•m²/s²
  • 211700 joule Answer

4 0
2 years ago
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