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sdas [7]
3 years ago
8

The U. S. Navy is helping these Congolese soldiers conduct an inspection of vehicles, looking mostly for car bombs. They can ins

pect areas under the car using a __________ and light's property of _________.
A) flat mirror; reflection
B) flat mirror; refraction
C) convex mirror; absorption
D) concave mirror; reflection
Physics
2 answers:
Mashutka [201]3 years ago
6 0
A flat mirror; reflection
Fed [463]3 years ago
5 0
The answer is a) Flat mirror; reflection.

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This chemical equation represents a ______________ reaction.
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you didnt put anything

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The spring to launch a pinball in a pinball machine is compressed 25 cm and has a spring constant of 140 N/m.
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8 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
3 years ago
A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug
vlada-n [284]

Answer:

The Minimum wavelength is  \lambda_{min}= 382.2nm

The Maximum wavelength is \lambda_{max}= 624.2nm

Explanation:

From the question we are told that  

              The energy range is  E_r = 3.25eV \ and  \ 19.9eV

   Considering E = 19.9eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 19.9 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{max}}

Here h is the Planck's constant with value of  h= 6.625*10^{-34}J\cdot s

        c is the speed of light with value of  c = 3*10^8 m/s

Substituting values and making \lambda the subject of the formula

                       \lambda_{max} = \frac{hc}{E}

                         = \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}

                         \lambda_{max}= 624.2nm

  Considering E = 3.25eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 3.25 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{min}}

Substituting values and making \lambda the subject of the formula

                       \lambda_{min} = \frac{hc}{E}

                           = \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}

                           \lambda_{min}= 382.2nm

3 0
3 years ago
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