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2 years ago

If you are swimming upstream (i.e., against the current), at what speed does your friend on the shore see you moving?

Physics

1 answer:

Alex777 [14]2 years ago

7 0

Could be very slow since they’re basically going against the current which is hard so will be going slow

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The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,

KatRina [158]

Answer:

Explanation:

I took the test

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1 year ago

The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20

NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

= 3.1536 x 10⁷ +0.840

= 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

= 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

= 86400.0023 s

The time period of rotation is given by the formula,

ωₐ = 2π / Tₐ

Substituting the values,

ωₐ = 2π / 365.046306

= 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

= 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

= 2π /86400

= 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration

α = (ωₓ - ωₐ) / T₁

= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²

3 0

2 years ago

What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an

Sindrei [870]

Answer:

$a = 12.78 g's$

Explanation:

Height reached by the object after push off is given as

$H = 2.3 m$

$v_f^2 - v_i^2 = 2 a s$

now we have

$0 - v^2 = 2(-9.81)(2.3)$

$v = 6.72 m/s$

now we know that this push last for total distance of 0.18 m

so during the push we will have

$v_f^2 - v_i^2 = 2 a d$

$6.72^2 - 0 = 2a(0.18)$

$a = 125.35 m/s^2$

now in terms of g = 9.81 m/s/s we have

$a = \frac{125.35}{9.81} gs$

$a = 12.78 g's$

6 0

1 year ago

Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light

Tpy6a [65]

Answer:

that technician A is right

Explanation:

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From the above it is seen that technician A is right

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2 years ago

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Zina [86]

The 1st,3rd,4th, and 5th ones are the correct answers

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2 years ago

Read 2 more answers