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2 years ago

An object is dropped from a height h. during the final second of its fall, it traverses a distance of 41.2 m. what was h?

1 answer:

3 0

41.2 = h-1/2g(t-1)^2
 {-h = -1/2gt^2-1/2g+g*t-41.2
 {h = 1/2gt^2
 summing them up
 0 = -1/2g+g*t-41.2
 41.2 +4.9 = g*t
 t = 46.1/9.8 = 4.70 sec
 h = 1/2gt^2 =4.9*(4.70^2) = 108.241 m

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you nose out another runner to win 100.000 m dash. if your total time for the race was 13.800 s and you aced out the other runne

sammy [17]

Your average speed was

(100 m) / (13.8 s) = 7.25 m/s .

If you finished 0.001s ahead of him, then at your average speed, that corresponds to

(7.25 m/s) x (0.001 s) = 0.00725 m

That's 7.25 millimeters ... about 0.28 of an inch !

NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.

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3 years ago

What is the second law of thermodynamics

andriy [413]

It states that the total entropy of an isolated system can never decrease over time

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2 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

2 years ago

You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find

ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

ω = 1.57 rad/s

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

ac = 4.92 m/s²

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

μs = 0.5

7 0

2 years ago

Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe

Ede4ka [16]

The final velocity ( $v_1_f$ ) of the first astronaut will be greater than the final velocity of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

The given parameters;

Mass of the first astronaut, = m₁
Mass of the second astronaut, = m₂
Initial velocity of the first astronaut, = v₁
Initial velocity of the second astronaut, = v₂ > v₁
Mass of the ball, = m
Speed of the ball, = u
Final velocity of the first astronaut, = $v_f_1$
Final velocity of the second astronaut, = $v_f_2$

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

$m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f$

if v₂ > v₁, then $v_1_f > v_2_f$ , to conserve the linear momentum.

Thus, the final velocity ( $v_1_f$ ) of the first astronaut will be greater than the final velocity of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

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2 years ago