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sergeinik [125]
4 years ago
8

Distance between bholu and golu house house 9 km . bholu has to attend golu birthday party at 7am . he started from his home at

6 am on his bicycle and covered a distance of 6 km in 40 mins . At that point he met chintu and he spoke to him for 5 mins and reached goku's birthday party at 7 am . with what speed did he cover the second part of the journey ? calculate his average speed for the entire journey?
Physics
2 answers:
navik [9.2K]4 years ago
6 0
For the rest 3 km


Bholu need to run in 5 minutes so


He need to 3 × 1000 / 5 × 60 m/s

3000/ 300 = 10 m/s


So , his speed should be 1 m/s to reach in 5 minute



mark as brainliest 5 star
bezimeni [28]4 years ago
6 0

Answer:

r mooooommkk/ljjkbtnnnnnbnbvvhhhgggggggjjjjjjjjjjkkjjjkbhbhghghfjhthdrsbrgghbfjkf

Explanation:

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Answer:

4.6×10^-7 m or 0.46nm

Explanation:

From

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At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
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Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

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g = w^2R

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3 years ago
A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of
Drupady [299]

Answer:

4.5\times 10^{-5} T

Explanation:

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Current in wire=40 A

Magnetic field=B_1=3.5\times 10^{-5} T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

B_{wire}=B_2=\frac{\mu_0I}{2\pi R}

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B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T

The resultant magnetic field is given by

B=\sqrt{B^2_1+B^2_2}

Substitute the values then we get

B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}

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The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire=4.5\times 10^{-5} T

Hence, the resultant magnitude of the magnetic field 29 cm above  the wire=4.5\times 10^{-5} T

7 0
3 years ago
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