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aleksklad [387]
2 years ago
10

The earth rotates about its pole once every 24 hrs. The distance from the pole to a location on the Earth is 35* north latitude

is about 3243.8 miles. Therefore, a location on Earth at 35* north latitude is spinning on a circle of radius 3242.8 miles. Compute the linear speed on the surface of the Earth at 35* north latitude.
Physics
1 answer:
Oksana_A [137]2 years ago
4 0
The angular velocity, ω= 
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s
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A body travels the first half of the total distance with velocity v and second half with v2 calculate avg velocity
LuckyWell [14K]

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v = 2 v₁ v₂ / (v₁ + v₂)

Explanation:

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8 0
2 years ago
Q1: A cyclist brakes to a stop. His thinking distance was 1m and his braking distance was 3m. What was his overall stopping dist
weeeeeb [17]

Answer:

1.) 4m

2.) 37 m

3.) 62m

4.) 2.5 s

Explanation:

1.) Given that the

Thinking distance = 1m

Breaking distance = 3m

Stopping distance = breaking distance + thinking distance

Stopping distance = 1 + 3 = 4m

2.) Given that the

Stopping distance = 52 m

Thinking distance = 15m

Breaking distance = 52 - 15 = 37m

3.) The stopping distance = 76m

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Breaking distance = 76 - 14 = 62m

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4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?

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5 0
3 years ago
A 2-kg toy car accelerates from 5 to 10 m/s.
andre [41]
<h3><u>Full Question:</u></h3>

A 2-kg toy car accelerates from 5 to 10 m/s in 2 seconds. Find the applied force.

<h2><u>Answer:</u></h2>

Force applied is 5\frac {kgm} {s^2}.

<h3><u>Explanation:</u></h3>

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The initial velocity of the body = 5 m/s.

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Mass of body = 2 kg.

So, force applied =  2 \times 2.5 kgm/s^2. =5kgm/s^2.

4 0
3 years ago
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