All categories

4 years ago

Which projectiles will be visibly affected by air resistance when they fall?

2 answers:

8 0

All of the above will be affected by air resistance, but the most obvious will be the balloon or leaf.
Hope it helps somewhat!

lozanna [386]4 years ago

5 0

Answer:

Balloon and leaf

Explanation:

You might be interested in

I need help on both a and b of question 1

marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

$r = \frac{\Delta c}{\Delta t}$

Given the following reaction:

$2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)$

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

$r = -\frac{\Delta [N_2O_5]}{2 \Delta t}$

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

$r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}$

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

$r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s$

(b) Using the relationship derived previously, we know that:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

Rate of appearance of nitrogen dioxide is given by:

$r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}$

Which is obtained from the equation:

$-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}$

If we multiply both sides by 4, that is:

$-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}$

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

5 0

3 years ago

Which of these is an empirical formula?

iVinArrow [24]

Answer: C

Explanation:

7 0

3 years ago

Write the beta decay equation for the following isotope: 91 38 Sr? Please write out steps

babymother [125]

Answer:

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}Y$

Explanation:

The unbalanced nuclear equation is

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e \, + \, ?$

Let's write the question mark as a nuclear symbol.

$\rm _{38}^{91}Se} \longrightarrow \, _{-1}^{0}e \, + \, _{Z}^{A}X$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Then

98 = 0 + A, so A = 98 - 0 = 98, and

38 = -1 + Z, so Z = 38 + 1 = 39

Then, your nuclear equation becomes

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}X$

Element 39 is yttrium, so the balanced nuclear equation is

$\rm _{38}^{91}Sr \longrightarrow \, _{-1}^{0}e + \, _{39}^{91}Y$

7 0

3 years ago

numbers represented by the following prefi xes: (a) mega-, (b) kilo-, (c) deci-, (d) centi-, (e) milli-, (f) micro-, (g) nano-,

Dahasolnce [82]

Explanation:

mega=10raised to the power 6

kilo=10 raised to the power 3

centi = 10 raised to the power negative 2

Milli = 10 raised to the power negative 3

nano = 10 raised to the power negative 9

pico= 10 raised to the power negative 12

micro = 10 raised to the power negative 6

5 0

3 years ago

What kind of bonding would exist between<br> sulfur and hydrogen?

Alexandra [31]

Answer:

nonpolar covalent

Explanation:

Comparing their electronegativities will help determine the type of bond. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The difference in electronegativity (ΔEN) is used to determine bond type. Hydrogen has an electronegativity of 2.20 and sulfur has 2.58. The difference is 0.38,so the electrons are shared in a nonpolar covalent bond.

5 0

3 years ago