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zheka24 [161]
3 years ago
5

A large centrifuge is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and at

mospheric reentries. At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation
Physics
1 answer:
melomori [17]3 years ago
6 0

Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

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An astronaut is on the moon. He drops a hammer from a height of 3.2metres and it takes 2.0 seconds to reach the lunar landscape.
Anvisha [2.4K]

Answer:

1/6 m/s^2      ( about 1/6th gravity of Earth ( 9.81 m/s^2)

Explanation:

Displacement =  yo  +  vo t  - 1/2 a t^2

      -  3.2          = 0     +  0     - 1/2 a(2.0)^2

      -     3.2       =                -2a

             a = 3.2 / 2 = 1.6 m/s^2

6 0
2 years ago
A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?
Lyrx [107]
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.

To calculate acceleration, divide your weight by the mass, thus the accleration is:

100N/100kg = 1(m/s^2)
7 0
3 years ago
Read 2 more answers
The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known
Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

m_1 = Mass of electron = 9.11\times 10^{−31}\ kg

v_1 = Speed of electron = 5.7\times 10^7\ m/s

p_2 = Neutrino has a momentum = 7.3\times 10^{-24}\ kg m/s

M = total mass = 2.34\times 10^{-26}\ kg

In the x axis as the momentum is conserved

Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s

In the y axis

Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s

The resultant velocity is

R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s

The recoil speed of the nucleus is 2240.92365 m/s

3 0
3 years ago
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0
3 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
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