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Bond [772]
3 years ago
10

A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image

Physics
1 answer:
lilavasa [31]3 years ago
6 0

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
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Coefficient of friction = 0.836

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If v be the speed after one quarter of the circular path

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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
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-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

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

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

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