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Helga [31]
3 years ago
7

Si un astronauta tiene una masa de 72kg si el valor de g en la luna es 1.6m/s2, ¿cual es la fuerza gravitacionar de la luna sobr

e el astronauta?​
Physics
1 answer:
Fantom [35]3 years ago
5 0

(72 kg) (1.6 m/s²) = 115.2 kg • m/s² = 115.2 N

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Electronegativity increases when atoms ___
notka56 [123]

your answer is.....

D. have a large atomic radius

although they also increase going from left to right so if D is incorrect, B might be your answer. it depends on context of the lesson.

8 0
3 years ago
When electrons are removed from the outermost shell of a calcium atom, the atom becomesA. an anion that has a larger radius than
Free_Kalibri [48]

Answer:

D. a cation that has a smaller radius than the atom.

Explanation:

When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.

3 0
3 years ago
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Are zebra fish schooling fish?
Alex73 [517]

Answer:

they r schooling fish

Explanation:

they need to be kept in groups of 5.

5 0
3 years ago
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In the two-slit experiment, monochromatic light of wavelength 600 nm passes through a 19) pair of slits separated by 2.20 x 10-5
kumpel [21]

Explanation:

It is given that,

Wavelength of monochromatic light, \lambda=600\ nm=6\times 10^{-7}\ m

Slits separation, d=2.2\times 10^{-5}\ m

(a) We need to find the angle corresponding to the first bright fringe. For bright fringe the equation is given as :

d\ sin\theta=n\lambda, n = 1

\theta=sin^{-1}(\dfrac{\lambda}{d})

\theta=sin^{-1}(\dfrac{6\times 10^{-7}}{2.2\times 10^{-5}})

\theta=1.56^{\circ}

(b) We need to find the angle corresponding to the second dark fringe, n = 1

So, d\ sin\theta=(n+\dfrac{1}{2})\lambda

sin\theta=\dfrac{3\lambda}{2d}

\theta=sin^{-1}(\dfrac{3\lambda}{2d})

\theta=sin^{-1}(\dfrac{3\times 6\times 10^{-7}}{2\times 2.2\times 10^{-5}})

\theta=2.34^{\circ}

Hence, this is the required solution.

4 0
3 years ago
A force of 1200 N is applied to a drum of radius 80 cm which has a 12 kg mass, m1, attached by a cord
maksim [4K]

Answer:

1.022 x 103 N.m

Explanation:

Solution

Given:

The weight of the block of mass m₂ is :

w₂ = m₂*g

Where

w₂ = 39 x 9.8 = 382.2 N

Then,

The weight of the block of mass m₁

w₁= m₁*g;

so,

w₁ = 12 x 9.8 = 117.5 N

Thus,

The tension wrapped in cord on drum (80 cm) T₁ = F - w₁

Now,

T₁ = 1200 - 117.5

T₁ = 1082.5 N

The tension  wrapped in the cord on drum (41 cm) T₂ = w₂;

T₂ = 382.2 N

Hence,

We calculate net torque on the center of the drum:

The net torque = T₁ x 0.8 + T₂ x 0.41;

= 1082.5 x 0.8 + 382.2 x 0.41;

= 1.022 x 103 N.m

Therefore, the resulting torque applied to the system is 1.022 x 103 N.m

4 0
3 years ago
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