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Ira Lisetskai [31]
3 years ago
8

Define moment of a force about a point and give the si unit of moment?​

Physics
1 answer:
tigry1 [53]3 years ago
5 0

Answer:

Moment of a force is the product of the force and the perpendicular distance of force from axis of rotation.

The SI unit of force is newton (N).

Explanation:

Hope it helps you again!

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Why are the trends for electronegativity and ionization energy similar
Goshia [24]

Answer:

Explanation:

Ionization Energy Trends

Ionization energy is the energy required to remove an electron from a neutral atom in its gaseous phase. Conceptually, ionization energy is the opposite of electronegativity. ... As a result, it is easier for valence shell electrons to ionize, and thus the ionization energy decreases down a group

3 0
3 years ago
Which has more electron shells: oxygen or sulfur? How do you know?
Sever21 [200]
Deffinitly oxygen cause everyone breathes and yah.
3 0
3 years ago
Read 2 more answers
A double slit that is illuminated with coherent light of wavelength 644 nm produces a pattern of bright and dark fringes on a sc
shutvik [7]

Answer:

2.77 cm

Explanation:

d = separation between the slits = 2783 x 10⁻⁹ m

\lambda = wavelength of coherent light = 644 nm = 644 x 10⁻⁹ m

D = Distance of the screen = 6 cm = 0.06 m

y_{n} = Position of nth bright fringe

Position of nth bright fringe is given as

y_{n} = \frac{nD\lambda }{d}  

for n = 2

y_{2} = \frac{nD\lambda }{d}  

y_{2} = \frac{(2)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}

y_{2} = 0.0278 m

for n = 4

y_{4} = \frac{nD\lambda }{d}  

y_{4} = \frac{(4)(0.06)(644\times10^{-9}))}{2783\times10^{-9}}

y_{4} = 0.0555 m

Distance between 4th and 2nd bright fringes is given as

w = y_{4} - y_{2} = 0.0555 - 0.0278 = 0.0277 m

w = 2.77 cm

8 0
3 years ago
The noble gases have eight valence electrons and as a result are
AURORKA [14]
The noble gases have eight valence electrons and as a result are stable. 

If an atom consists of 8 valence electrons, they have a full octet, and do not need to bond, which makes them "happy".
4 0
4 years ago
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
3 years ago
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