Ask question

Ask question

All categories

4 years ago

10

You run from your friends house that is 1k away. You then walk home. What distance did you travel?

1 answer:

kolezko [41]4 years ago

6 0

Would would have moved a total distance of 2k, because walking is a slower speed does not change distance

You might be interested in

I'm kind of getting bored. Someone please talk to me.

kumpel [21]

Answer:

Sure um how is your day going

Explanation:

7 0

3 years ago

Read 2 more answers

An airplane is traveling due east with a velocity of 7.5 × 102 kilometers/hour. There is a tailwind of 30 kilometers/hour. What

Delvig [45]

A tailwind means you add thed velocity of the tailwind to the velovity of the plane (no need for any vectors etc. so you just need (7.5x 102 ) +30 then convert it so it like one of the answers (btw - none of the answers is exactly correct - you need to round to get it) thats what you need to do to get the question

4 0

3 years ago

Read 2 more answers

Based on the graph, describe how momentum changes with time for an object in free fall. If you can help, I would be so grateful.

pochemuha

With time, momentum increases as it builds speed assuming their is nothing in the way to stop it. Based on the graph, you can see that example being displayed as the line on the graph gets higher

4 0

3 years ago

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe

JulijaS [17]

Answer:

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.$

Explanation:

Given charges are:

$\rm q_1 = +6e.\\q_2 = -6e.$

The electric potential energy of a charge due to the electric field of another charge is given by

$\rm EPE=\dfrac{kq_1q_2}{r}.$

where,

k = Coulomb's constant, having value = $\rm 9\times 10^9\ Nm^2/C^2.$
r = distance between the charges.

When the charges are infinite distance apart, $\rm r = \infty$ ,

$\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.$

When the charges are $\rm 5.61\times 10^{-12}\ m$ apart, $\rm r=5.61\times 10^{-12}\ m$ ,

$\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.$

Here, e is the charge on one electron, such that, $\rm e = -1.6\times 10^{-19}\ C$ .

Therefore,

$\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.$

Thus,

$\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.$

4 0

3 years ago

A chemist needs 15.0 grams of ethanol for a reaction. If the density of ethanol is 0.789 g/ml, how many milliliters of alcohol s

harina [27]

Mass=15g
Density=0.789g/ml

$\\ \bull\tt\dashrightarrow Density=\dfrac{Mass}{Volume}$

$\\ \bull\tt\dashrightarrow Volume=\dfrac{Mass}{Density}$

$\\ \bull\tt\dashrightarrow Volume=\dfrac{15}{0.789}$

$\\ \bull\tt\dashrightarrow Volume=19.01mL$

3 0

3 years ago