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bazaltina [42]
3 years ago
9

Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right.” Is Lorenzo cor

rect? If so,why? Explain his error
Physics
2 answers:
ziro4ka [17]3 years ago
8 0

Sample Response: Lorenzo is not correct. Nonmetals increase in reactivity from left to right because nonmetals on the right have more valence electrons. They need to gain fewer electrons to have a full outer shell. However, this trend only continues until group 17, because the noble gases already have a full outer shell. Therefore, their reactivity is the least of all the elements.

~theLocoCoco

Rzqust [24]3 years ago
6 0

Answer:

No, he is not correct. Xenon is not very reactive since it is located in the group of noble gases. The noble gases are all chemically unreactive gases.Elements within different groups within the periodic table have different physical and chemical properties. Hope this answers the question.

Explanation:

Yes

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Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo
Tems11 [23]

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

3 0
4 years ago
What is significant figures​
Maurinko [17]

Explanation:

Significant figures of a number written in positional notation are digits that carry meaningful contributions to the measurement resolution of the number.

5 0
4 years ago
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a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 m/s the centripetal force on t
jek_recluse [69]
Coupla things wrong with this question, Sam.
Let's clean those up first, and then we'll work on the answer.

-- The car is NOT moving with uniform velocity.
'Velocity' includes both speed and direction.  If either of these
changes, it's a change of velocity.
On a circular track, the car's direction is CONSTANTLY changing,
so its velocity is too. 
The thing that's uniform is its speed, not its velocity.

-- A 'neutron' is a subatomic particle found in the nucleus of most
atoms.  It's not a unit of force.  The unit of force is the 'Newton'.
_______________________

OK. A centripetal force of 6,000 newtons keeps 1,200 kg of mass
moving in a circle at 20 m/s.

The formula:

                                     Centripetal force = (mass) (speed)² / (radius)

Multiply each side
by 'radius':                (centripetal force) x (radius) = (mass) x (speed)²

Divide each side by
'centripetal force':       Radius = (mass) x (speed)² / (centripetal force)

Write in the numbers
that we know:              Radius = (1200 kg) (20 m/s)² / (6000 Newtons)

                                                 = (1200 kg) (400 m²/s²) / (6000 Newtons)

                                                 = (480,000 kg-m²/s²) / (6000 kg-m/s²)

                                                 = (480,000 / 6000) meters
                                                
                                                 =         80 meters .   
8 0
3 years ago
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Answer:

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4 0
3 years ago
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A particle passes through the point P=(−3,1,0)P=(−3,1,0) at time t=3t=3, moving with constant velocity v⃗ =⟨5,3,−2⟩v→=⟨5,3,−2⟩.
eimsori [14]

Answer:

The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

Explanation:

Given that,

The point is

P=(-3,1,0)

Time t = 3

Velocity v=(5,3,-2)

We need to calculate the parametric equation for the position of the particle

Using parametric equation for position

r(t)=r_{0}+v(t)t....(I)

at t = 3,

P=r(t)

Put the value into the formula

(-3,1,0)=r_{0}+(5,3,-2)\times3

(-3,1,0)=r_{0}+(15,9,-6)

r_{0}=(-18,-8,6)

Put the value of r₀ in equation (I)

r(t)=(-18,-8,6)+(5,3,-2)t

r(t)=(-18+5t,-8+3t,6-2t)

Hence, The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

3 0
3 years ago
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