All categories

3 years ago

Find the velocity at which the bowling ball must move for its kinetic energy to be equal to that of the meteorite.

Physics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

Explanation:

BOOM!!!

You might be interested in

How much energy will a stock tank heater rated at 1790 Watts use in a 24 hour period? 1. 1790 × 24 × 3600 Joules 2. 1790 × 24 ×

Rashid [163]

<h2>Option 1 is the correct answer.</h2>

Explanation:

Power of heater, P = 1790 W

Time used, t = 24 hours = 24 x 60 x 60 = 24 x 3600 s

We have the equation

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}$

We need to find energy,

Substituting

$\texttt{Power}=\frac{\texttt{Energy}}{\texttt{Time}}\\\\1790=\frac{\texttt{Energy}}{24\times 3600}$

Energy = 1790 x 24 x 3600 J

Option 1 is the correct answer.

3 0

3 years ago

Please Please Help!!, What happens when a current moves through a wire and the wire is placed near a magnet?

USPshnik [31]

I think it is c not sure

7 0

3 years ago

Read 2 more answers

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago

Which of the following quantities are unknown?

Sauron [17]

Answer:

Charge of the alpha particle

Explanation:

6 0

3 years ago

Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

$x = rcos\theta$

$y = rsin\theta$

for r = 30 sec

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

7 0

3 years ago