Natalie accelerates her skateboard along a straight path from 4 m/s to 0 m/s in 25.0 s. Find the acceleration.

What do solutions and colloids have in common

Andreyy89

I believe thye answer is either d or c

6 0

3 years ago

Read 2 more answers

PLEASE HELP ME IM ON A TIMER

garri49 [273]

Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

$A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j$

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

$Displacement = \sqrt{(3)^2+ (6)^2} \\Displacement = 6.70$

And the angle will be defined by:

tan(beta)=3/6

beta = tan^-1(6/3)

beta = 63.43°

5 0

3 years ago

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

IrinaK [193]

Answer:

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

Explanation:

We are given;

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

Converting to seconds, we have;

T₁ = (365 × 24 × 60 × 60) + 0.84

T₁ = (3.1536 x 10⁷) + 0.840

T₁ = 31536000.84 s

Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days

Converting to seconds, we have;

T₀ = 31536000 s

Hence, time period of one rotation in the year 2006 is;

Tₐ = 31536000.84/365

Tₐ = 86400.0023 s

The time period of rotation is given by the formula;

Tₐ = 2π/ωₐ

Making ωₐ the subject;

ωₐ = 2π/Tₐ

Plugging in the relevant values;

ωₐ = 2π/ 365.046306

ωₐ = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation in the year 1906 is;

Tₓ = 31536000/365

Tₓ = 86400 s

Time period of rotation,

Tₓ = 2π /ωₓ

ωₓ = 2π / T

Plugging in the relevant values;

ωₓ = 2π/86400

ωₓ = 7.272205217 x 10⁻⁵ rad/s

The average angular acceleration is given by;

α = (ωₓ - ωₐ) / T₁

α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84

α = 6.152 X 10⁻²⁰ rad/s²

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

8 0

4 years ago

What is the definition of the word amplitude

qwelly [4]

Explanation:

Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path. ... Waves are generated by vibrating sources, their amplitude being proportional to the amplitude of the source.

8 0

3 years ago

A simple pendulum is 1.15 meters long and has a period of 6.29 seconds. The pendulum is on an unknown planet. What is the accele

Nastasia [14]

Answer:

The value of the acceleration of gravity of the Unknown Planet = 1.14 $\frac{m}{s^{2} }$