Ask question

Ask question

All categories

3 years ago

8

Newton's first law of motion states that an object remains at rest unless a(n) ____ force acts on it. a. balanced c. gravitation

al b. frictional d. unbalanced Please select the best answer from the choice

2 answers:

Sauron [17]3 years ago

8 0

It is an unbalanced force.

Rufina [12.5K]3 years ago

7 0

It is an unbalanced force.

You might be interested in

A skateboarder is standing at the top of a tall ramp waiting to begin a trip. The skateboarder has

Mnenie [13.5K]

Has a skateboard. your gonna have to give more details the. that just one .

7 0

3 years ago

Are you country or are you city?

Mariulka [41]

Answer:

Neither lma0 I'm from a town :P

Explanation:

Hbu?

Have a nice dayyy <3

6 0

3 years ago

Read 2 more answers

If the mass of the moon is ...

bulgar [2K]

C decreased the factor cuz the max is smaller

8 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 $1.00$ $atm$ , 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

4 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

Olin [163]

The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: $\mu m g$ , where $\mu=0.25$ is the coefficient of friction and $m=30.0 kg$ is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
$F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N$
And so, this is the force that the worker must apply to the crate.

5 0

3 years ago