34.1 grams / 1.1 mL =

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

3 years ago

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

3 years ago

What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan

kramer

I think the question should be the below:

What is the total distance, side to side, that the top of the building moves during such an oscillation?

Answer is the below:

Acceleration .. a = (-) ω² x 
(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) 

x (max) = a(max) /ω² 

x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)

5 0

3 years ago

The action force is the balloon pushing the air out. What is the magnitude of the reaction force of the air pushing on the ballo

aleksley [76]

Understanding Newtons second law that everything has an equal and opposite reaction. The reaction force from a balloons air being pushed out is the preasurized air it had to push out into the open air.

7 0

3 years ago

Read 2 more answers

Explain why more stars are circumpolar for observers at higher latitudes.

PIT_PIT [208]

Explanation:

When you observe the night sky you will notice that the stars are moving. They rise from eastern horizon and set in the western horizon. It happens due to rotation of Earth. When observed closely you will notice that the all the stars seem to go around the pole star. Out of all the stars there are some stars which neither set not rise, such stars are called as Circumpolar stars. This means that they are always above the horizon. If we trace the path of such stars they will appear to make complete circle around the pole star.

Also, you will notice that the altitude of pole star (separation of pole star from the horizon in degrees) will depend on the location of observe on the Earth. This happens due to Earth being spherical. So if you are on equator the pole star will be on the horizon i.e. 0° altitude. If you are at Poles, altitude of the pole star will be 90°. Technically the altitude of pole star at any place on Earth is equal to the latitude of the place.

If the altitude of pole star varies and increases as you move towards higher latitude on Earth, the distance between horizon and pole star will also increase. This will result in more stars being circumpolar.

If you are at Poles, all the stars will be circumpolar and if you are at equator no star will be circumpolar.

8 0

3 years ago