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3 years ago

Will the waves be reflected with inversion or without? how do you know?

Physics

1 answer:

anygoal [31]3 years ago

3 0

Inversion will be your answer

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X + 10 times X = 50 what is the answer

tia_tia [17]

Answer:

4.54

Explanation:

X+10X=50

11X=50

X=4.54#

<h2>please follow me...</h2>

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3 years ago

While performing an experiment involving a reaction between two chemicals, the scientist observes that the reaction container ha

AVprozaik [17]

When an object is becomes warm its an example of exothermic reaction.

8 0

3 years ago

An egg is dropped from a building that is 61 m high.

Allisa [31]

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

Initial Velocity = 0 m/s

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

t = 3.5 s

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

3 0

3 years ago

If 128 v is required to push 4.00 A of current though a resistor, what is the resistance

prisoha [69]

Answer:

32.0 Ω

Explanation:

Ohm's law:

V = IR

128 V = (4.00 A) R

R = 32.0 Ω

8 0

3 years ago

Be sure to answer all parts. Without stratospheric ozone (O3), harmful solar radiation would cause gene alterations. Ozone forms

Ratling [72]

Answer:

for C-Cl bond $\lambda_1=3.66\times 10^{-9}\ m$

for O-O bond $\lambda_2=2.43\times 10^{-9}\ m$

Explanation:

We have the bond enegy for C-Cl, $BE_1=327\ kJ.mol^{-1}= 327000\times 1.66\times 10^{-22}\ J\ per\ molecule$

bond enegy for C-Cl, $BE_2=494\ kJ.mol^{-1}= 494000\times 1.66\times 10^{-22}\ J\ per\ molecule$

Now as we know the energy of electromagnetic waves is given by:

$E=h.\nu$

here E = BE

Then for C-Cl:

$327000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_1$

$\nu_1=8.18733\times 10^{16}\ Hz$

Now wavelength:

$\lambda_1=\frac{c}{\nu_1}$

$\lambda_1=\frac{3\times 10^8}{8.18733\times 10^{16}}$

$\lambda_1=3.66\times 10^{-9}\ m$

For O2:

$494000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_2$

$\nu_2=1.236863\times 10^{17}\ Hz$

Now wavelength:

$\lambda_2=\frac{c}{\nu_2}$

$\lambda_2=\frac{3\times 10^8}{1.236863\times 10^{17}}$

$\lambda_2=2.43\times 10^{-9}\ m$

8 0

3 years ago