Ask question

Ask question

All categories

2 years ago

14

An arrow is launched with an initial velocity of 7.6 m/s at an angle of 42 degrees above the horizontal. What is the horizontal

component of the velocity at the highest point of the trajectory?

1 answer:

AleksAgata [21]2 years ago

5 0

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

$v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}$

$\:\:\:\:\:=5.6\:\text{m/s}$

You might be interested in

solve the following system by any method 2x - 6y = 24 ... -5x + 6y = -6 ... A. (0,-6) B. (4,-1) C. (-6,-6) D. (6,-1)

vodka [1.7K]

Using elimination, the answer is C.

5 0

3 years ago

Why do you think football players are usually large in size?

kiruha [24]

Football players in the NFL are large in size due to their eating habits since as players they have to work out constantly to stay and shape and eat to gain weight and be strong, so its important for them to be strong against their opponents in gameplay. Football players sometimes take steroids which make their muscles stronger and harder, and also gives them a growth spurt but the effects are worse and makes them sicker, high blood pressure, and even more.

8 0

3 years ago

Read 2 more answers

Sirens and flashing lights do NOT indicate that

sergij07 [2.7K]

Golf cart is the answer

7 0

3 years ago

Read 2 more answers

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

3 years ago

A coffee filter of mass 1.2 grams dropped from a height of 1 m reaches the ground with a speed of 0.8 m/s. How much kinetic ener

iren [92.7K]

<u>Answer:</u> The energy gained by the air molecules is 0.011 J.

<u>Explanation:</u>

Law of conservation of energy states that energy can neither be created nor be destroyed but it can only be transformed from one form to another form.

Here, the potential energy of the coffee filter is getting converted into kinetic energy of the coffee filter and some energy is lost by it which is gained by the air molecules in the form of kinetic energy.

So, calculating the potential energy of coffee filter, we use the equation:

P = mgh

where,

m = mass of coffee filter = 1.2 g = $1.2\times 10^{-3}kg$ (Conversion used: 1 kg = 1000 g)

g = acceleration due to gravity = $9.8m/s^2$

h = height of coffee filter = 1 m

Putting values in above equation, we get:

$P=1.2\times 10^{-3}kg\times 9.8m/s^2\times 1m\\\\P=1.176\times 10^{-2}J$

Calculating the kinetic energy of coffee filter, we use the equation:

$E=\frac{1}{2}mv^2$

where,

m = mass of coffee filter = 1.2 g = $1.2\times 10^{-3}kg$

v = speed of coffee filter = 0.8 m/s

Putting values in above equation, we get:

$E=\frac{1}{2}\times 1.2\times 10^{-3}kg\times (0.8m/s)^2\\\\E=3.84\times 10^{-4}J$

As, energy lost by coffee filter = energy gained by air molecules

So, energy lost by coffee filter = Potential energy - Kinetic energy

Energy lost by coffee filter = $(1.176\times 10^{-2})-(3.84\times 10^{-4})=0.011J$

Hence, the energy gained by the air molecules is 0.011 J.

5 0

3 years ago