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riadik2000 [5.3K]
3 years ago
10

You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distan

ce between the two of you remains the same, so you deduce that you are walking at the same speed of 1.09 m/s . Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.85 m/s . It takes you 14.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running?
Physics
1 answer:
Setler79 [48]3 years ago
5 0

Answer:

25.52 m

Explanation:

Relative speed between the person and I would be the difference of our speeds

v_r=2.85-1.09=1.76\ m/s

Time taken by me to walk up to the person = 14.5 s

Distance is given by

Distance=Speed\times Time\\\Rightarrow s=vt\\\Rightarrow s=1.76\times 14.5\\\Rightarrow s=25.52\ m

The person was 25.52 m ahead of me when I started running

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Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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The aeroplane lands at a speed of 80 m/s
strojnjashka [21]

The mass of the aeroplane is 300,000 kg.

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The law is represented as

F =ma

where acceleration a = velocity change v / time interval t

Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.

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Substitute the values and we have

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Thus, the mass of the aeroplane is 300,000 kg.

Learn more about Newton's second law of motion.

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8 0
2 years ago
Plzpzlpzlzplzplzplzpz this one also<br> all questions ​
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Answer:

I didn't know these questions sorry

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