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Dennis_Churaev [7]

3 years ago

9

A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi

ne its resultant displacement vector by use of the diagram.

1 answer:

KatRina [158]3 years ago

3 0

Answer:

Explanation is given

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!!!!HURRY 30 POINTS!!!!

Nataly_w [17]

The anwser is C flooding

3 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

A +6.33 uC charge q1 is attracted by a force of 0.115 N to a second charge q2 that is 1.44 m away. What is the value of q2?

jarptica [38.1K]

Answer:-4.1858

Explanation: I hate Acellus

7 0

2 years ago

In a classroom demonstration, the pressure inside a soft drink can is suddenly reduced to essentially zero. Assuming the can to

umka2103 [35]

Answer:

3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.

Explanation:

The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =

A = 2πr(r + h)

Where h = height of the can = 12cm

r = radius of the can = 6.5cm/2 = 3.25cm

r = diameter /2

A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²

Atmospheric pressure, P = 101325Pa = 101325 N/m²

F = P × A

F = 101325 ×0.031.

F = 3141N. Or 3.1 ×10³ N.

5 0

3 years ago

______ involves organizing and breaking down information into easier groups to expand capacity. Rehearsal is the verbal repetiti

adell [148]

Hi!

The answers would be <u>chunking</u> & <u>short-term</u>

1. <u>Chunking </u>involves organizing and breaking down information into easier groups to expand capacity.

<h3>Explanation:</h3>

Chunking is a mental process that is observed to increase short-term memory by taking the information and categorizing it into small groups. For instance, a longer number taken as a single unit is harder to recall then when it is divided into smaller units. 235469350 is harder to instantly recall as compared to when it is chunked into 3 groups: 235 469 350.

This allows more information to be stored in, thereby increasing the capacity of the mind to store information.

2. Rehearsal is the verbal repetition of information. These techniques are especially important for the improvement of <u>short-term</u> memory.

<h3>Explanation: </h3>

Short-term memory is lost after a couple of seconds or minutes, for instance even if you chunk the information, you might not recall it after 30 seconds. Rehearsing or repetition of information, either loudly or mentally, extends the time a particular information is retained.

So you depending on the number of times you repeat the number 235 469 350, the more your short term memory improves .

Hope this helps!

6 0

3 years ago