:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?

A 1234 kg freight car moving at 6 m/s runs into a 2468 kg freight car at rest. They stick together upon collision. What was the

kirill [66]

Answer:

2 m/s

Explanation:

Applying,

The law of conservation of momentum

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the first freight car, m' = mass of the second freight car, u = initial velocity of the first freight car, u' = initial velocity of the second freight car, V = final combined velocity/ speed.

make V the subject of the equation

V = (mu+m'u')/(m+m')........... Equation 2

From the question,

Given: m = 1234 kg, m' = 2468 kg, u = 6 m/s, u' = 0 m/s (at rest)

Substitute these values into equation 2

V = [(1234×6)+(2468×0)]/(1234+2468)

V = 7404/3702

V = 2 m/s

5 0

3 years ago

A Scooter with a mass of 50 kilograms travels with a speed of 10 meters per second. A 2 Nt force bring it to a halt. How long ti

Serga [27]

Answer:

250s

Explanation:

Given parameters:

Mass = 50kg

Speed = 10m/s

Force = 2N

Unknown:

Time the force was acting = ?

Solution:

To solve this problem, we use the equation of motion from the second law of Newton:

Ft = mV

F is the force

t is the time taken

m is the mass

V is the velocity

2 x t = 50 x 10

t = 250s

3 0

3 years ago

Assume that a satellite orbits mars 150km above its surface. Given that the mass of mars is 6.485 X 10^23kg, and the radius of m

Kisachek [45]

<span>3598 seconds The orbital period of a satellite is u=GM p = sqrt((4*pi/u)*a^3) Where p = period u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits. a = semi-major axis of orbit. Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2 The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So 150000 m + 3.396x10^6 m = 3.546x10^6 m Substitute the known values into the equation for the period. So p = sqrt((4 * pi / u) * a^3) p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3) p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3) p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3) p = sqrt(1.2945785x10^7 s^2) p = 3598.025212 s Rounding to 4 significant figures, gives us 3598 seconds.</span>

8 0

3 years ago

HELP + extra pts // Two 10-m high diving platforms are at opposing ends of a 30-m pool. How fast must two clowns run straight of

ZanzabumX [31]

Explanation:

The clowns need to leave the diving boards with enough horizontal velocity such that each travels 15 m (half the width of the pool) in the same time that they fall (vertically) the 10-m from the top of the diving board.

We'll assume no force acts on the clowns horizontally to slow them down while they are in flight. And we'll assume that only gravity acts on the clowns vertically.

We can treat the horizontal and vertical components separately. This will help simplify the problem.

Let's start with the vertical displacement. Let's say the clown is dropped from a height of 10-m. How long would it take them to fall that distance?

Using our equations of motion (with constant, linear acceleration), we can solve for this time. $d = vt + \frac{1}{2} at^2$

Where d is the distance travelled, v is the initial velocity, a is acceleration, and t is time.

If the clown is dropped, they have an initial velocity of 0 (zero). We assumed only gravity acts on the clown, so acceleration equals the gravitational acceleration on earth. $a = 9.8m/s^2$

The distance the clown travels is 10-m from the diving board to the surface of the water.

Let's solve. $10 = 0*t + \frac{1}{2} (9.8) t^2 \rightarrow 20/9.8 = t^2 \rightarrow t = \sqrt{20/9.8}$

Now that we know time, we can calculate how fast the clown needs to be running when they leap from the diving board to cover a distance of 15-m (Remember, half the width of the pool.)

Using our equations of motion, we know that $d = vt + 0.5at^2$

We assumed no forces act horizontally on the clown, therefore a = 0. We just need to solve for v. Substituting in the time we just solved for, we get something like this. $15 = v \sqrt{20/9.8}$

I'll leave it to you to solve this equation.

6 0

3 years ago

What is the name of the kind of stretch that involves stretching as far as you can and then holding for 10-30 seconds

Dmitriy789 [7]

Answer:

Static stretching.

Explanation:

It is static stretching because it is a form of stretching which u can do actively for a period of time and you hold position for about 30 to 60 seconds which allow the muscles and connective tissues to lengthen. It is done after work out with out movement in order to loosen up muscles so as to gain flexibility.

7 0

3 years ago

:) What is practical machine? what is the redation between MA and VR in a practicalmachine?​

:) What is practical machine? what is the redation between MA and VR in a practicalmachine?