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neonofarm [45]
2 years ago
15

A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such

devices because they can simulate the same Accelerations that are experienced in certain types of flight. At which angular velocit would a space station of 30 m radius have to rotate to simulate Earth graivty?
Physics
1 answer:
Leviafan [203]2 years ago
5 0

Answer:

ω = 0.571 rad/s

Explanation:

given data

radius = 30 m

solution

we take here g = 9.8 m/s²

and g is express as

g = r × ω²     ....................1

put here value and we get

9.8 = 30 × ω²

solve it we get

ω = 0.571 rad/s

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An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist
Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0
3 years ago
What matches ????????????????
dangina [55]

Answer:

1st: Radiation

2nd: Conduction

3rd: Convection

Explanation:

I actually learned this before in school. Yay

8 0
3 years ago
A fireperson is 50 m from a burning building and directs a stream of water from a fire hose at an angle of 300 above the horizon
notsponge [240]

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

5 0
3 years ago
Find the energy in joules required to lift a 55.0 megagram object a distance of 500cm
pantera1 [17]

1,000 grams = 1 kilogram
so 55 megagrams = 55,000 kilograms

100 cm = 1 meter
so 500 cm = 5 meters

Acceleration of gravity on Earth = 9.8 m/s²

Weight = (mass) x (gravity)

========================================

Work = increase in potential energy =

               (weight) x (height) =

             (mass) x (gravity) x (height) =

             (55,000 kg) x (9.8 m/s²) x (5 m) =

                     2,695,000 joules .

5 0
3 years ago
A 0.5-kilogram ball is thrown vertically upward with an initial kinetic energy of 25 joules. Ap-proximately how high will the ba
tatuchka [14]
It's gravitational potential energy at the top will roughly equal it's kinetic energy when it was released (a little is lost to air resistance).  Note this will assume the release point is zero potential energy.  (we are free to define it that way, just letting you know).  Gravitational potential energy is mgh.
mgh=25J
h=25J/(0.5kg x 9.81m/s^2) = 5.097m
So it goes about 5.1 meters above the point where it was released
4 0
3 years ago
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