All categories

3 years ago

If a car's initial velocity is 24 m/s E, and 3 seconds later the car's

Physics

1 answer:

gizmo_the_mogwai [7]3 years ago

4 0

Please find the attached photo for your answer. Hope it helps..

You might be interested in

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 Potential Energy

grandymaker [24]

Answer:

energy that is ponteaila

Explanation:

8 0

3 years ago

Read 2 more answers

A total charge of Q coulomb is uniformly distributed along a rod 40cm in length.find find the electric field intensity 20cm away

Tatiana [17]

For a total charge of Q coulomb is uniformly distributed along a rod 40cm in length, the electric field intensity 20cm away from the rod is mathematically given as

E1=1.598*10^11v/m

<h3>What is the electric field intensity 20cm away from the rod along its perpendicular bisector?</h3>

Generally, the equation for the initial electric field intensity is mathematically given as

$dEp=\int{kd/r}cosd\theta$

Therefore

$e1=kd/r{sin\theta2+sinR1}$

Hence

$E1=(9*10^9/20*10^{-2})({sin45+sin45})*B/40*10^-2$

E1=B*9*10^{13})/(10*110)* $\sqrt{2}$

E1=1.598*10^11v/m

In conclusion, the electric field intensity

E1=1.598*10^11v/m