A bus is moving with an acceleration of 4 m/s^2. If it was initially at rest,

A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does

lorasvet [3.4K]

Answer:

A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.

Explanation:

4 0

3 years ago

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

Artist 52 [7]

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s ................1

we know acceleration will be

a = $\frac{-u^2}{2\times S}$

Then we have

Force = mass × acceleration .................2

m × $\frac{-u^2}{2\times S}$ = -μk mg

The puck decelerates due to the kinetic frictional force μk mg

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

4 0

2 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

3 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

2 years ago

The moon does not have atmosphere. Why?

Galina-37 [17]

Answer: It doesn't because it's not that big.

Explanation:

6 0

3 years ago

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