A basketball weighting 0.5kg is flying through the air at 10m/s. what is the kinetic energy

An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.

ivann1987 [24]

<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters

</span>

5 0

3 years ago

A stone is thrown vertically upwards with a speed of 30.0 m/s.

matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s

b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m

6 0

3 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

2 years ago

Your Lead Teaching Assistant is initially seated on the top of a hemispherical ice mound of radius R = 30 m. Approximate the ice

saw5 [17]

Answer:

<em>Height = 5.65 km</em>

Explanation:

$3\pi r^2$ is the circumference or we can say measures the boundary of hemisphere of friction-less ice that he is sitting on.

So, the height will be = 2 x 3.14 x (30)^2 = 5654.7 m = 5.65 km

7 0

2 years ago

Read 2 more answers

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago