If you had 6,000 milligrams of rock to move how many kilograms is that?

irakobra [83]

Answer:177.87

Explanation:

4 0

3 years ago

A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1

attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.

Explanation:

Given:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

To Find:

The average force and momentum

Formulas:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

Solution:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.

5 0

3 years ago

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1250 kg, which is required to travel upward 49 m in

insens350 [35]

Answer:

659.01W

Explanation:

The cab has a mass of 1250 kg, the weight of the cab represented by Wc will be

Wc = mass of the cab × acceleration due to gravity in m/s²

Wc = 1250 × 9.81 = 12262.5 N

but the counter weight of the elevator represented by We = mass × acceleration due to gravity = 995 × 9.81 = 9760.95 N

Net weight = weight of the cab - counter weight of the elevator = Wc - We = 12262.5 - 9760.95 = 2501.55 N

the motor of the elevator will have to provide this in form of work

work done by the elevator to lift the cab to height of 49 m = net weight × distance (height) = 2501.55 × 49m

power provided by the motor of the elevator = workdone by the motor / time in seconds

Power = (2501.55 × 49) ÷ ( 3.1 × 60 seconds) = 659.01 W

5 0

3 years ago

The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e

Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ( $v$ ), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

$v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi$ (1)

Where:

$\Delta t$ - Rotation time, measured in seconds.

$R$ - Radius of the Earth, measured in meters.

$\phi$ - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that $\Delta t = 86160\,s$ , $R = 6.371\times 10^{6}\,m$ and $\phi = 37.6819^{\circ}$ , then the tangential speed at Livermore is:

$v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}$

$v\approx 284.001\,\frac{m}{s}$

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0

3 years ago

A larger object is made of two balls separated by a massless rigid rod that is 1.5 m long. The mass of the red ball at one end i

lions [1.4K]

Answer:

The speed of the 1 kg red ball 8.04 m/s .

Explanation:

Given :

Separation between rods , d = 1.5 m .

Mass of the red ball is 1 kg .

Mass of the orange ball is 5.7 kg .

Angular velocity , $\omega=1\ rev/s = 2\pi\ rad/s$ .

Now , distance of center of mass from red ball is :

$r =\dfrac{0\times 1+1.5\times 5.7}{1+5.7}\\\\r = 1.28\ m$

We know , speed is given by :

$v=\omega r\\\\v=2 \times \pi \times 1.28\\\\v=8.04\ m/s$

Hence , this is the required solution .

5 0

3 years ago