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S_A_V [24]
3 years ago
15

Imagine a cart moving with constant velocity over a piece of paper. If the cart makes a mark directly beneath it on the paper at

a regular time interval, in this case every 0.1 s, the marks would look like diagram A above. This is known as a “ticker tape” diagram of motion.
In the ticker tape diagram above, where does the cart have positive acceleration?
Where does it have negative acceleration?
Explain what features in diagram B you interpret as positive and negative acceleration.
If the dots were spaced widely apart in the diagram, then what could you infer about the speed of the cart?
Physics
1 answer:
spin [16.1K]3 years ago
8 0

Answer:

wait... are harley from 4th grade?

cuz itz meh haley

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

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Explanation:

hope this is right.

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