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Kaylis [27]
3 years ago
11

Freshly picked cucumbers are dropped into a bin from a height of 1.25 m above the bottom of the bin. Assuming that the bin is em

pty, how fast is a cucumber going when it hits the bottom of the bin? How much time does it take the cucumber to fall to the bottom of the bin?
Physics
1 answer:
postnew [5]3 years ago
8 0

Answer:

a) Vf = 4.95 m/s

b) t = 0.51s

Explanation:

take downwards as positive.

let Vf be the final velocity as the cucumber reach the bottom of the bin and Vi be the initial velocity of the cucumber when they dropped.

a ) from equations of motion:

(Vf)^2 = (vi)^2 +2g×(x - x0) ,

<em>since x0 = 0 m and Vi = 0 m/s.</em>

Vf = \sqrt{2g×(x)}  = \sqrt{2(9.8)×(1.25)} = 4.95 m/s, downwards

therefore, the cucumber will reach the bottom of the bin with a speed of 4.95 m/s.

b) from equations of motion:

Vf = Vi + g×t

Vi = 0 then:

t = Vf/g = 4.95/9.8 = 0.51 s

therefore, the cucumber will take 0.51 seconds to reach the bottom of the bin.

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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.

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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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At what time of year is the intensity of solar radiation striking each of earth's hemispheres weakest?
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The resultant velocity is given by

v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

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