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Step2247 [10]
3 years ago
12

What is the recurvate react

Physics
1 answer:
deff fn [24]3 years ago
3 0
Bent or curved backwords

You might be interested in
How do you answer this. Need help ASAP. Offering 20 points !!!
Temka [501]

in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.

your temp should be converted in kelvin

variables:

p1=3.0×10^6 n/m^2

t1= 270k

just add 273 to your celcius

p2= ? your solving for this

t2= 315k

then you set up the equation

(3.0×10^6)/270= (x)(315)

you then cross multiply

(3.0×10^6)315=270x

distribute the 315 to the pressure.

9.45×10^8=270x then you divide 270 o both sides to get

answer

3.5×10^6 n/m^2

7 0
3 years ago
The vertical displacement of the wave is measured from the ?
sergey [27]
The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.

Equilibrium to the crest . . . that's the amplitude.

Crest to trough . . . that's double the amplitude.

Trough to trough . . . How did that get in here ?  Yes, that's
                               the wavelength, but it has nothing to do
                               with vertical displacement.

Frequency . . . that's how many complete waves pass a mark
                       on the ground every second.  Doesn't belong here.

Notice that this has to be a transverse wave.  If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.

It also has to be a vertically 'polarized' wave.  If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement.  It still has an amplitude, but the amplitude
is all horizontal.
6 0
3 years ago
A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0
3 years ago
The person given credit for the first modern atomic theory was __________.
mote1985 [20]
<span>The person given credit for the first modern atomic theory was __________.


Dalton.</span>
7 0
3 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
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